Introduction to the theory of divided differences

Tiberiu Popoviciu
Approximation Theory, divided differences, Numerical Analysis

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Tiberiu Popoviciu

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T. Popoviciu, Introduction à la théorie des différences divisées, Bull. Math. de la Soc. Roum. des Sci., t. 42 (1940), pp. 65-78 (in French)

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1940 a -Popoviciu- Bull. Math. Soc. Roum. Sci. – Introduction a la theorie des differences divisees

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Bulletin mathématique de la Société Roumaine des Sciences

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Societatea de Științe Matematice din România

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https://www.jstor.org/stable/43769841

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[MR0013171,  Zbl 0025.39803,JFM 66.0389.01].

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1940 a -Popoviciu- Bull. Math. Soc. Roum. Sci. - Introduction a la theorie des differences divisees.
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INTRODUCTION TO THE THEORY OF DIVIDED DIFFERENCES

BY

TIBERIU POPOVICIU

In what follows we propose to give the demonstration of several formulas which intervene in the theory of divided differences of functions of one variable. Most of these formulas are already exposed, without demonstration, in our Thesis 1 1 ^(1){ }^{1}1), These formulas being in common use in the theory of higher order convex functions, it will not be useless to establish them here in all rigor. I decided to return to these questions by noting the appearance of some works where my results are not cited 2 2 ^(2){ }^{2}2).
  1. The Lagrange polynomial. Consider n + 1 n + 1 n+1n+1n+1distinct points x 1 , x 2 , … , x n + 1 x 1 , x 2 , … , x n + 1 x_(1),x_(2),dots,x_(n+1)x_{1}, x_{2}, \ldots, x_{n+1}x1,x2,…,xn+1and either f = f ( x ) f = f ( x ) f=f(x)f=f(x)f=f(x)a uniform function defined on these points. We will assume that the points x i x i x_(i)x_{i}xiare on the real axis and that f f fffis a real function.
There exists one and only one polynomial of degree n which takes the values f ( x i ) f x i f(x_(i))\mathrm{f}\left(\mathrm{x}_{i}\right)f(xi)to the points x i , i = 1 , 2 , … , n + 1 3 ) x i , i = 1 , 2 , … , n + 1 3 {:x_(i),i=1,2,dots,n+1^(3))\left.\mathrm{x}_{i}, \mathrm{i}=1,2, \ldots, \mathrm{n}+1{ }^{3}\right)xi,i=1,2,…,n+13).
The existence of at least one polynomial satisfying the imposed conditions can be demonstrated by induction. For n = 0 n = 0 n=0n=0n=0the property is immediate. The constant f ( x 1 ) f x 1 f(x_(1))f\left(x_{1}\right)f(x1)(polynomial of degree 0) satisfies the imposed condition. Suppose that the property is true for n n nnnand we will demonstrate it for n + 1 n + 1 n+1n+1n+1. There therefore exists, by hypothesis, at least one polynomial P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)of degree n − 1 n − 1 n-1n-1n−1taking the values f ( x i ) f x i f(x_(i))f\left(x_{i}\right)f(xi)to the points x i , i = 1 x i , i = 1 x_(i),i=1x_{i}, i=1xi,i=1, 2 , … , n 2 , … , n 2,dots,n2, \ldots, n2,…,n. We then see that the polynomial
P ( x ) + [ f ( x n + 1 ) − P ( x n + 1 ) ] ( x − x 1 ) ( x − x 2 ) … ( x − x n ) ( x n + 1 − x 1 ) ( x n + 1 − x 2 ) … ( x n + 1 − x n ) P ( x ) + f x n + 1 − P x n + 1 x − x 1 x − x 2 … x − x n x n + 1 − x 1 x n + 1 − x 2 … x n + 1 − x n P(x)+[f(x_(n+1))-P(x_(n+1))]((x-x_(1))(x-x_(2))dots(x-x_(n)))/((x_(n+1)-x_(1))(x_(n+1)-x_(2))dots(x_(n+1)-x_(n)))P(x)+\left[f\left(x_{n+1}\right)-P\left(x_{n+1}\right)\right] \frac{\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{n}\right)}{\left(x_{n+1}-x_{1}\right)\left(x_{n+1}-x_{2}\right) \ldots\left(x_{n+1}-x_{n}\right)}P(x)+[f(xn+1)−P(xn+1)](x−x1)(x−x2)…(x−xn)(xn+1−x1)(xn+1−x2)…(xn+1−xn)
is of degree n n nnnand takes the values f ( x i ) f x i f(x_(i))f\left(x_{i}\right)f(xi)to the points x i , i = 1 , 2 , … , n + 1 x i , i = 1 , 2 , … , n + 1 x_(i),i=1,2,dots,n+1x_{i}, i=1,2, \ldots, n+1xi,i=1,2,…,n+1.
Uniqueness is easily obtained. If we had two polynomials P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)And Q ( x ) Q ( x ) Q(x)\mathrm{Q}(x)Q(x), not identical, of degree n n nnnand verifying the imposed conditions, the difference P ( x ) − Q ( x ) P ( x ) − Q ( x ) P(x)-Q(x)\mathrm{P}(x)-\mathrm{Q}(x)P(x)−Q(x)would be zero for x = x 1 , x 2 , … , x n + 1 x = x 1 , x 2 , … , x n + 1 x=x_(1),x_(2),dots,x_(n+1)x=x_{1}, x_{2}, \ldots, x_{n+1}x=x1,x2,…,xn+1, which is impossible.
The unique polynomial, determined as we have seen, is called the Lagrange (interpolation) polynomial of the function f f fffon the points x i , i = 1 , 2 , … , n + 1 x i , i = 1 , 2 , … , n + 1 x_(i),i=1,2,dots,n+1\mathrm{x}_{i}, \mathrm{i}=1,2, \ldots, \mathrm{n}+1xi,i=1,2,…,n+1.
We will designate this polynomial by
(1) P ( x 1 , x 2 , … , x n + 1 ; f ∣ x ) . (1) P x 1 , x 2 , … , x n + 1 ; f ∣ x . {:(1)P(x_(1),x_(2),dots,x_(n+1);f∣x).:}\begin{equation*} \mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f \mid x\right) . \tag{1} \end{equation*}(1)P(x1,x2,…,xn+1;f∣x).
The general form of polynomials taking the values f ( x i ) f x i f(x_(i))f\left(x_{i}\right)f(xi)to the points x i , i = 1 , 2 , … , n + 1 x i , i = 1 , 2 , … , n + 1 x_(i),i=1,2,dots,n+1x_{i}, i=1,2, \ldots, n+1xi,i=1,2,…,n+1, East
P ( x 1 , x 2 , … , x n + 1 ; f ∣ x ) + ( x − x 1 ) ( x − x 2 ) … ( x − x n + 1 ) Q ( x ) P x 1 , x 2 , … , x n + 1 ; f ∣ x + x − x 1 x − x 2 … x − x n + 1 Q ( x ) P(x_(1),x_(2),dots,x_(n+1);f∣x)+(x-x_(1))(x-x_(2))dots(x-x_(n+1))Q(x)\mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f \mid x\right)+\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{n+1}\right) \mathrm{Q}(x)P(x1,x2,…,xn+1;f∣x)+(x−x1)(x−x2)…(x−xn+1)Q(x)
Q ( x ) Q ( x ) Q(x)\mathrm{Q}(x)Q(x)being any polynomial 4 4 ^(4){ }^{4}4).
The uniqueness of the polynomial (1) allows us to write the following well-known formulas,
P ( x 1 , x 2 , … , x n + 1 ; f ∣ x ) = P x 1 , x 2 , … , x n + 1 ; f ∣ x = P(x_(1),x_(2),dots,x_(n+1);f∣x)=\mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f \mid x\right)=P(x1,x2,…,xn+1;f∣x)=
= ∑ i = 1 n + 1 f ( x i ) ( x − x 1 ) … ( x − x i − 1 ) ( x − x i + 1 ) … ( x − x n + 1 ) ( x i − x 1 ) … ( x i − x i − 1 ) ( x i − x i + 1 ) … ( x i − x n + 1 ) = ∑ i = 1 n + 1 f ( x i ) ? ( x ) P ′ ( x i ) ( x − x i ) = ∑ i = 1 n + 1   f x i x − x 1 … x − x i − 1 x − x i + 1 … x − x n + 1 x i − x 1 … x i − x i − 1 x i − x i + 1 … x i − x n + 1 = ∑ i = 1 n + 1   f x i ? ( x ) P ′ x i x − x i =sum_(i=1)^(n+1)f(x_(i))((x-x_(1))dots(x-x_(i-1))(x-x_(i+1))dots(x-x_(n+1)))/((x_(i)-x_(1))dots(x_(i)-x_(i-1))(x_(i)-x_(i+1))dots(x_(i)-x_(n+1)))=sum_(i=1)^(n+1)(f(x_(i))?(x))/(P^(')(x_(i))(x-x_(i)))=\sum_{i=1}^{n+1} f\left(x_{i}\right) \frac{\left(x-x_{1}\right) \ldots\left(x-x_{i-1}\right)\left(x-x_{i+1}\right) \ldots\left(x-x_{n+1}\right)}{\left(x_{i}-x_{1}\right) \ldots\left(x_{i}-x_{i-1}\right)\left(x_{i}-x_{i+1}\right) \ldots\left(x_{i}-x_{n+1}\right)}=\sum_{i=1}^{n+1} \frac{f\left(x_{i}\right) ?(x)}{\mathcal{P}^{\prime}\left(x_{i}\right)\left(x-x_{i}\right)}=∑i=1n+1f(xi)(x−x1)…(x−xi−1)(x−xi+1)…(x−xn+1)(xi−x1)…(xi−xi−1)(xi−xi+1)…(xi−xn+1)=∑i=1n+1f(xi)?(x)P′(xi)(x−xi),
P ( x 1 , x 2 , … , x n + 1 ; f ∣ x ) = − | 1 x 1 x 1 2 … x 1 n f ( x 1 ) 1 x 2 x 2 2 … x 2 n f ( x 2 ) … … … … … … … 1 x n + 1 x n + 1 2 … x n + 1 n f ( x n + 1 ) 1 x x 2 … x n 0 | | 1 x 1 x 1 2 … x 1 n 1 x 2 x 2 2 … x 2 n … … … … … … 1 x n + 1 x n + 1 2 … x n + 1 n | P x 1 , x 2 , … , x n + 1 ; f ∣ x = − 1 x 1 x 1 2 … x 1 n f x 1 1 x 2 x 2 2 … x 2 n f x 2 … … … … … … … 1 x n + 1 x n + 1 2 … x n + 1 n f x n + 1 1 x x 2 … x n 0 1 x 1 x 1 2 … x 1 n 1 x 2 x 2 2 … x 2 n … … … … … … 1 x n + 1 x n + 1 2 … x n + 1 n P(x_(1),x_(2),dots,x_(n+1);f∣x)=-(|[1,x_(1),x_(1)^(2),dots,x_(1)^(n),f(x_(1))],[1,x_(2),x_(2)^(2),dots,x_(2)^(n),f(x_(2))],[dots dots,dots,dots,dots,dots,dots],[1,x_(n+1),x_(n+1)^(2),dots,x_(n+1)^(n),f(x_(n+1))],[1,x,x^(2),dots,x^(n),0]|)/(|[1,x_(1),x_(1)^(2),dots,x_(1)^(n)],[1,x_(2),x_(2)^(2),dots,x_(2)^(n)],[dots dots dots,dots,dots,dots],[1,x_(n+1),x_(n+1)^(2),dots,x_(n+1)^(n)]|)\mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f \mid x\right)=-\frac{\left|\begin{array}{cccccc} 1 & x_{1} & x_{1}^{2} & \ldots & x_{1}^{n} & f\left(x_{1}\right) \\ 1 & x_{2} & x_{2}^{2} & \ldots & x_{2}^{n} & f\left(x_{2}\right) \\ \ldots \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 1 & x_{n+1} & x_{n+1}^{2} & \ldots & x_{n+1}^{n} & f\left(x_{n+1}\right) \\ 1 & x & x^{2} & \ldots & x^{n} & 0 \end{array}\right|}{\left|\begin{array}{ccccc} 1 & x_{1} & x_{1}^{2} & \ldots & x_{1}^{n} \\ 1 & x_{2} & x_{2}^{2} & \ldots & x_{2}^{n} \\ \ldots \ldots \ldots & \ldots & \ldots & \ldots \\ 1 & x_{n+1} & x_{n+1}^{2} & \ldots & x_{n+1}^{n} \end{array}\right|}P(x1,x2,…,xn+1;f∣x)=−|1x1x12…x1nf(x1)1x2x22…x2nf(x2)…………………1xn+1xn+12…xn+1nf(xn+1)1xx2…xn0||1x1x12…x1n1x2x22…x2n………………1xn+1xn+12…xn+1n|
where we posed
(2) f ( x ) = ( x − x 1 ) ( x − x 2 ) … ( x − x n + 1 ) . (2) f ( x ) = x − x 1 x − x 2 … x − x n + 1 . {:(2)f(x)=(x-x_(1))(x-x_(2))dots(x-x_(n+1)).:}\begin{equation*} f(x)=\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{n+1}\right) . \tag{2} \end{equation*}(2)f(x)=(x−x1)(x−x2)…(x−xn+1).
Let's pose
(3)
[ x 1 , x 2 , … , x n + 1 ; f ] = ∑ i = 1 n + 1 f ( x i ) P ′ ( x i ) x 1 , x 2 , … , x n + 1 ; f = ∑ i = 1 n + 1   f x i P ′ x i [x_(1),x_(2),dots,x_(n+1);f]=sum_(i=1)^(n+1)(f(x_(i)))/(P^(')(x_(i)))\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right]=\sum_{i=1}^{n+1} \frac{f\left(x_{i}\right)}{\mathcal{P}^{\prime}\left(x_{i}\right)}[x1,x2,…,xn+1;f]=∑i=1n+1f(xi)P′(xi)
4 4 ^(4){ }^{4}4) The polynomial (1) can also be characterized by the following extremal property. There exists one and only one polynomial of minimum effective degree which takes the values f ( x i ) f x i f(x_(i))f\left(x_{i}\right)f(xi), For x = x i , i = 1 , 2 , … , n + 1 x = x i , i = 1 , 2 , … , n + 1 x=x_(i),i=1,2,dots,n+1x=x_{i}, i=1,2, \ldots, n+1x=xi,i=1,2,…,n+1.
which is the coefficient of x n x n x^(n)x^{n}xnin the polynomial (1). This polynomial can then be written in the following forms 5 5 ^(5){ }^{5}5)
P ( x 1 , x 2 , … , x n ; f ∣ x ) + [ x 1 , x 2 , … , x n + 1 ; f ] φ ( x ) x − x n + 1 P ( x 2 , x 3 , … , x n + 1 ; f ∣ x ) + [ x 1 , x 2 , … , x n + 1 ; f ] φ ( x ) x − x 1 P x 1 , x 2 , … , x n ; f ∣ x + x 1 , x 2 , … , x n + 1 ; f φ ( x ) x − x n + 1 P x 2 , x 3 , … , x n + 1 ; f ∣ x + x 1 , x 2 , … , x n + 1 ; f φ ( x ) x − x 1 {:[P(x_(1),x_(2),dots,x_(n);f∣x)+[x_(1),x_(2),dots,x_(n+1);f](varphi(x))/(x-x_(n+1))],[P(x_(2),x_(3),dots,x_(n+1);f∣x)+[x_(1),x_(2),dots,x_(n+1);f](varphi(x))/(x-x_(1))]:}\begin{aligned} & \mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n} ; f \mid x\right)+\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right] \frac{\varphi(x)}{x-x_{n+1}} \\ & \mathrm{P}\left(x_{2}, x_{3}, \ldots, x_{n+1} ; f \mid x\right)+\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right] \frac{\varphi(x)}{x-x_{1}} \end{aligned}P(x1,x2,…,xn;f∣x)+[x1,x2,…,xn+1;f]φ(x)x−xn+1P(x2,x3,…,xn+1;f∣x)+[x1,x2,…,xn+1;f]φ(x)x−x1
from which results the recurrence relation of Lagrange polynomials
(4) P ( x 1 , x 2 , … , x n + 1 ; f ∣ x ) = = ( x − x 1 ) P ( x 2 , x 3 , … , x n + 1 ; f ∣ x ) − ( x − x n + 1 ) P ( x 1 , x 2 , … , x n ; f ∣ x ) x n + 1 − x 1 (4) P x 1 , x 2 , … , x n + 1 ; f ∣ x = = x − x 1 P x 2 , x 3 , … , x n + 1 ; f ∣ x − x − x n + 1 P x 1 , x 2 , … , x n ; f ∣ x x n + 1 − x 1 {:[(4)P(x_(1),x_(2),dots,x_(n+1);f∣x)=],[=((x-x_(1))P(x_(2),x_(3),dots,x_(n+1);f∣x)-(x-x_(n+1))P(x_(1),x_(2),dots,x_(n);f∣x))/(x_(n+1)-x_(1))]:}\begin{gather*} \mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f \mid x\right)= \tag{4}\\ =\frac{\left(x-x_{1}\right) \mathrm{P}\left(x_{2}, x_{3}, \ldots, x_{n+1} ; f \mid x\right)-\left(x-x_{n+1}\right) \mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n} ; f \mid x\right)}{x_{n+1}-x_{1}} \end{gather*}(4)P(x1,x2,…,xn+1;f∣x)==(x−x1)P(x2,x3,…,xn+1;f∣x)−(x−xn+1)P(x1,x2,…,xn;f∣x)xn+1−x1
  1. Divided differences. The divided difference of order n of the function f on the points x 1 , x 2 , … , x n + 1 x 1 , x 2 , … , x n + 1 x_(1),x_(2),dots,x_(n+1)\mathrm{x}_{1}, \mathrm{x}_{2}, \ldots, \mathrm{x}_{n+1}x1,x2,…,xn+1is characterized by the following properties:
    a) It is a linear and homogeneous expression with respect to f ( x 1 ) , f ( x 2 ) , … , f ( x n + 1 ) f x 1 , f x 2 , … , f x n + 1 f(x_(1)),f(x_(2)),dots,f(x_(n+1))\mathrm{f}\left(\mathrm{x}_{1}\right), \mathrm{f}\left(\mathrm{x}_{2}\right), \ldots, \mathrm{f}\left(\mathrm{x}_{n+1}\right)f(x1),f(x2),…,f(xn+1), with coefficients independent of the function f .
    b) It is zero identically for the functions f = 1 , x , x 2 , … f = 1 , x , x 2 , … f=1,x,x^(2),dots\mathrm{f}=1, \mathrm{x}, \mathrm{x}^{2}, \ldotsf=1,x,x2,…, x n + 1 x n + 1 x^(n+1)x^{n+1}xn+1.
    c) It reduces to 1 identically for the function f = x n f = x n f=x^(n)\mathrm{f}=\mathrm{x}^{n}f=xn.
The expression determined by these conditions a a aahas), b b bbb), c c ccc), is unique since the system
∑ i = 1 n + 1 λ i x i m = { 0 , m = 0 , 1 , … , n − 1 , 1 , m = n , ∑ i = 1 n + 1   λ i x i m = 0 ,      m = 0 , 1 , … , n − 1 , 1 ,      m = n , sum_(i=1)^(n+1)lambda_(i)x_(i)^(m)={[0",",m=0","1","dots","n-1","],[1",",m=n","]:}\sum_{i=1}^{n+1} \lambda_{i} x_{i}^{m}= \begin{cases}0, & m=0,1, \ldots, n-1, \\ 1, & m=n,\end{cases}∑i=1n+1λixim={0,m=0,1,…,n−1,1,m=n,
is a system of C Ramer C Ramer  C_("Ramer ")\mathrm{C}_{\text {Ramer }}CRow in λ 1 , λ 2 , … , λ n + 1 λ 1 , λ 2 , … , λ n + 1 lambda_(1),lambda_(2),dots,lambda_(n+1)\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n+1}λ1,λ2,…,λn+1The determinant of this system is, in fact, the Vandermonde determinant
V ( x 1 , x 2 … , x n + 1 ) = ∏ i < j ( x j − x i ) V x 1 , x 2 … , x n + 1 = ∏ i < j   x j − x i V(x_(1),x_(2)dots,x_(n+1))=prod_(i < j)(x_(j)-x_(i))\mathrm{V}\left(x_{1}, x_{2} \ldots, x_{n+1}\right)=\prod_{i<j}\left(x_{j}-x_{i}\right)V(x1,x2…,xn+1)=∏i<I(xI−xi)
distinct numbers x i x i x_(i)x_{i}xi.
Note that expression (3) verifies the properties a a aahas), b b bbb), c c ccc) since
P ( x 1 , x 2 , … , x n + 1 ; x i ∣ x ) = x i , i = 0 , 1 , … , n − 1 , P x 1 , x 2 , … , x n + 1 ; x i ∣ x = x i , i = 0 , 1 , … , n − 1 , P(x_(1),x_(2),dots,x_(n+1);x^(i)∣x)=x^(i),i=0,1,dots,n-1,\mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; x^{i} \mid x\right)=x^{i}, i=0,1, \ldots, n-1,P(x1,x2,…,xn+1;xi∣x)=xi,i=0,1,…,n−1,
it is therefore the value of the divided difference of f on the points x l x l x_(l)\mathrm{x}_{l}xL.
Let us designate by U ( x 1 , x 2 , … , x n + 1 ; f ) U x 1 , x 2 , … , x n + 1 ; f U(x_(1),x_(2),dots,x_(n+1);f)\mathrm{U}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f\right)U(x1,x2,…,xn+1;f)the determinant that we obtain from V ( x 1 , x 2 , … , x n + 1 ) V x 1 , x 2 , … , x n + 1 V(x_(1),x_(2),dots,x_(n+1))V\left(x_{1}, x_{2}, \ldots, x_{n+1}\right)V(x1,x2,…,xn+1)if we replace the elements x i n x i n x_(i)^(n)x_{i}^{n}xinby f ( x i ) f x i f(x_(i))f\left(x_{i}\right)f(xi)respectively. The divided difference (3) is then written
(5) [ x 1 , x 2 , … , x n + 1 ; f ] = U ( x 1 , x 2 , … , x n + 1 ; f ) V ( x 1 , x 2 , … , x n + 1 ) (5) x 1 , x 2 , … , x n + 1 ; f = U x 1 , x 2 , … , x n + 1 ; f V x 1 , x 2 , … , x n + 1 {:(5)[x_(1),x_(2),dots,x_(n+1);f]=(U(x_(1),x_(2),dots,x_(n+1);f))/(V(x_(1),x_(2),dots,x_(n+1))):}\begin{equation*} \left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right]=\frac{\mathrm{U}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f\right)}{\mathrm{V}\left(x_{1}, x_{2}, \ldots, x_{n+1}\right)} \tag{5} \end{equation*}(5)[x1,x2,…,xn+1;f]=U(x1,x2,…,xn+1;f)V(x1,x2,…,xn+1)
in the form of the quotient of two determinants. We see that the divided difference (5) is symmetrical with respect to the points x i x i x_(i)\mathrm{x}_{i}xi.
Formula (4) gives us the recurrence formula for divided differences
(6) [ x 1 , x 2 , … , x n + 1 ; f ] = [ x 2 , x 3 , … , x n + 1 ; f ] − [ x 1 , x 2 , … , x n ; f ] x n + 1 − x 1 [ x 1 ; f ] = f ( x 1 ) (6) x 1 , x 2 , … , x n + 1 ; f = x 2 , x 3 , … , x n + 1 ; f − x 1 , x 2 , … , x n ; f x n + 1 − x 1 x 1 ; f = f x 1 {:[(6)[x_(1),x_(2),dots,x_(n+1);f]=([x_(2),x_(3),dots,x_(n+1);f]-[x_(1),x_(2),dots,x_(n);f])/(x_(n+1)-x_(1))],[[x_(1);f]=f(x_(1))]:}\begin{gather*} {\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right]=\frac{\left[x_{2}, x_{3}, \ldots, x_{n+1} ; f\right]-\left[x_{1}, x_{2}, \ldots, x_{n} ; f\right]}{x_{n+1}-x_{1}}} \tag{6}\\ {\left[x_{1} ; f\right]=f\left(x_{1}\right)} \end{gather*}(6)[x1,x2,…,xn+1;f]=[x2,x3,…,xn+1;f]−[x1,x2,…,xn;f]xn+1−x1[x1;f]=f(x1)
which can also be used to define them and which justifies their name.
We can therefore write
(7) [ x 1 , x 2 , … , x n + 1 ; x m ] = { 0 , m = 0 , 1 , … , n − 1 1 , m = n (7) x 1 , x 2 , … , x n + 1 ; x m = 0 , m = 0 , 1 , … , n − 1 1 , m = n {:(7)[x_(1),x_(2),dots,x_(n+1);x^(m)]={[0",",m=0","1","dots","n-1],[1",",m=n]:}:}\left[x_{1}, x_{2}, \ldots, x_{n+1} ; x^{m}\right]= \begin{cases}0, & m=0,1, \ldots, n-1 \tag{7}\\ 1, & m=n\end{cases}(7)[x1,x2,…,xn+1;xm]={0,m=0,1,…,n−11,m=n
We have
(8) [ x 1 , x 2 , … , x n + 1 ; C f ] = C [ x 1 , x 2 , … , x n + 1 ; f ] (8) x 1 , x 2 , … , x n + 1 ; C f = C x 1 , x 2 , … , x n + 1 ; f {:(8)[x_(1),x_(2),dots,x_(n+1);Cf]=C[x_(1),x_(2),dots,x_(n+1);f]:}\begin{equation*} \left[x_{1}, x_{2}, \ldots, x_{n+1} ; \mathrm{C} f\right]=\mathrm{C}\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right] \tag{8} \end{equation*}(8)[x1,x2,…,xn+1;Cf]=C[x1,x2,…,xn+1;f]
(9) [ x 1 , x 2 … , x n + 1 ; f + g ] = [ x 1 , x 2 , … , x n + 1 ; f ] + [ x 1 , x 2 , … , x n + 1 ; g ] x 1 , x 2 … , x n + 1 ; f + g = x 1 , x 2 , … , x n + 1 ; f + x 1 , x 2 , … , x n + 1 ; g quad[x_(1),x_(2)dots,x_(n+1);f+g]=[x_(1),x_(2),dots,x_(n+1);f]+[x_(1),x_(2),dots,x_(n+1);g]\quad\left[x_{1}, x_{2} \ldots, x_{n+1} ; f+g\right]=\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right]+\left[x_{1}, x_{2}, \ldots, x_{n+1} ; g\right][x1,x2…,xn+1;f+g]=[x1,x2,…,xn+1;f]+[x1,x2,…,xn+1;g]
C being a constant and f , g f , g f,gf, gf,gtwo functions defined on the points x , i = 1 , 2 , … , n + 1 x ,  i = 1 , 2 , … , n + 1 x_(", ")i=1,2,dots,n+1x_{\text {, }} i=1,2, \ldots, n+1x, i=1,2,…,n+1.
If { f m } f m {f_(m)}\left\{f_{m}\right\}{fm}is a convergent sequence of functions defined on the points x i x i x_(i)x_{i}xi, we obviously have
lim m → ∞ [ x 1 , x 2 , … , x n + 1 ; f m ] = [ x 1 , x 2 , … , x n + 1 ; lim m → ∞ f m ] lim m → ∞   x 1 , x 2 , … , x n + 1 ; f m = x 1 , x 2 , … , x n + 1 ; lim m → ∞   f m lim_(m rarr oo)[x_(1),x_(2),dots,x_(n+1);f_(m)]=[x_(1),x_(2),dots,x_(n+1);lim_(m rarr oo)f_(m)]\lim _{m \rightarrow \infty}\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f_{m}\right]=\left[x_{1}, x_{2}, \ldots, x_{n+1} ; \lim _{m \rightarrow \infty} f_{m}\right]limm→∞[x1,x2,…,xn+1;fm]=[x1,x2,…,xn+1;limm→∞fm]
so, if the series ∑ m = 0 ∞ f m ∑ m = 0 ∞   f m sum_(m=0)^(oo)f_(m)\sum_{m=0}^{\infty} f_{m}∑m=0∞fmconverges, we have
∑ m = 0 ∞ [ x 1 , x 2 , … , x n + 1 ; f m ] = [ x 1 , x 2 … , x n + 1 ; ∑ m = 0 ∞ f m ] ∑ m = 0 ∞   x 1 , x 2 , … , x n + 1 ; f m = x 1 , x 2 … , x n + 1 ; ∑ m = 0 ∞   f m sum_(m=0)^(oo)[x_(1),x_(2),dots,x_(n+1);f_(m)]=[x_(1),x_(2)dots,x_(n+1);sum_(m=0)^(oo)f_(m)]\sum_{m=0}^{\infty}\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f_{m}\right]=\left[x_{1}, x_{2} \ldots, x_{n+1} ; \sum_{m=0}^{\infty} f_{m}\right]∑m=0∞[x1,x2,…,xn+1;fm]=[x1,x2…,xn+1;∑m=0∞fm]
From the above it follows that the divided difference of order n n nnnof a polynomial of degree n − 1 n − 1 n-1n-1n−1is still zero.
Let us note again the formula
f ( x ) − P ( x 1 , x 2 , … , x n + 1 ; f ∣ x ) = φ ( x ) [ x 1 , x 2 , … , x n + 1 , x ; f ] f ( x ) − P x 1 , x 2 , … , x n + 1 ; f ∣ x = φ ( x ) x 1 , x 2 , … , x n + 1 , x ; f f(x)-P(x_(1),x_(2),dots,x_(n+1);f∣x)=varphi(x)[x_(1),x_(2),dots,x_(n+1),x;f]f(x)-\mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f \mid x\right)=\varphi(x)\left[x_{1}, x_{2}, \ldots, x_{n+1}, x ; f\right]f(x)−P(x1,x2,…,xn+1;f∣x)=φ(x)[x1,x2,…,xn+1,x;f]
which is a simple relationship between the Lagrange polynomial and the divided difference.
Application. Let us calculate the divided differences (7) for m m mmmentire > n > n > n>n>n. Without restricting the generality we can assume x i ≠ 0 , i = 1 , 2 … x i ≠ 0 , i = 1 , 2 … x_(i)!=0,i=1,2dotsx_{i} \neq 0, i=1,2 \ldotsxi≠0,i=1,2…, n + 1 n + 1 n+1n+1n+1and either | z | < min ( 1 | x 1 | , 1 | x 2 | , … , 1 | x n + 1 | ) , z | z | < min 1 x 1 , 1 x 2 , … , 1 x n + 1 , z |z| < min((1)/(|x_(1)|),(1)/(|x_(2)|),dots,(1)/(|x_(n+1)|)),z|z|<\min \left(\frac{1}{\left|x_{1}\right|}, \frac{1}{\left|x_{2}\right|}, \ldots, \frac{1}{\left|x_{n+1}\right|}\right), z|z|<min(1|x1|,1|x2|,…,1|xn+1|),zbeing a parameter independent of x x xxx. We have
z n ( 1 − z x 1 ) ( 1 − z x 2 ) … ( 1 − z x n + 1 ) = ∑ i = 1 n + 1 1 ( 1 − z x i ) ′ o q ′ ( x i ) = = [ x 1 , x 2 , … , x n + 1 ; 1 1 − z x ] z n 1 − z x 1 1 − z x 2 … 1 − z x n + 1 = ∑ i = 1 n + 1   1 1 − z x i ′ o q ′ x i = = x 1 , x 2 , … , x n + 1 ; 1 1 − z x {:[(z^(n))/((1-zx_(1))(1-zx_(2))dots(1-zx_(n+1)))=sum_(i=1)^(n+1)(1)/((1-zx_(i))^(')(o)/(q)^(')(x_(i)))=],[=[x_(1),x_(2),dots,x_(n+1);(1)/(1-zx)]]:}\begin{gathered} \frac{z^{n}}{\left(1-z x_{1}\right)\left(1-z x_{2}\right) \ldots\left(1-z x_{n+1}\right)}=\sum_{i=1}^{n+1} \frac{1}{\left(1-z x_{i}\right)^{\prime} \frac{o}{q}^{\prime}\left(x_{i}\right)}= \\ =\left[x_{1}, x_{2}, \ldots, x_{n+1} ; \frac{1}{1-z x}\right] \end{gathered}zn(1−zx1)(1−zx2)…(1−zxn+1)=∑i=1n+11(1−zxi)′oq′(xi)==[x1,x2,…,xn+1;11−zx]
But,
1 1 − z x = z n x n 1 − z x + 1 + z x + z 2 x 2 + … + z n − 1 x n − 1 1 1 − z x = z n x n 1 − z x + 1 + z x + z 2 x 2 + … + z n − 1 x n − 1 (1)/(1-zx)=(z^(n)x^(n))/(1-zx)+1+zx+z^(2)x^(2)+dots+z^(n-1)x^(n-1)\frac{1}{1-z x}=\frac{z^{n} x^{n}}{1-z x}+1+z x+z^{2} x^{2}+\ldots+z^{n-1} x^{n-1}11−zx=znxn1−zx+1+zx+z2x2+…+zn−1xn−1
and taking into account (7), (8), (9),
[ x 1 , x 2 , … , x n + 1 ; 1 1 − z x ] = z n [ x 1 , x 2 , … , x n + 1 ; x n 1 − z x ] x 1 , x 2 , … , x n + 1 ; 1 1 − z x = z n x 1 , x 2 , … , x n + 1 ; x n 1 − z x [x_(1),x_(2),dots,x_(n+1);(1)/(1-zx)]=z^(n)[x_(1),x_(2),dots,x_(n+1);(x^(n))/(1-zx)]\left[x_{1}, x_{2}, \ldots, x_{n+1} ; \frac{1}{1-z x}\right]=z^{n}\left[x_{1}, x_{2}, \ldots, x_{n+1} ; \frac{x^{n}}{1-z x}\right][x1,x2,…,xn+1;11−zx]=zn[x1,x2,…,xn+1;xn1−zx]
If we therefore pose
(10) 1 ( 1 − z x 1 ) ( 1 − z x 2 ) … ( 1 ⋯ z x n + 1 ) = S 0 + S 1 z + S 2 z 2 + … 1 1 − z x 1 1 − z x 2 … 1 ⋯ z x n + 1 = S 0 + S 1 z + S 2 z 2 + … (1)/((1-zx_(1))(1-zx_(2))dots(1cdots zx_(n+1)))=S_(0)+S_(1)z+S_(2)z^(2)+dots\frac{1}{\left(1-z x_{1}\right)\left(1-z x_{2}\right) \ldots\left(1 \cdots z x_{n+1}\right)}=\mathrm{S}_{0}+\mathrm{S}_{1} z+\mathrm{S}_{2} z^{2}+\ldots1(1−zx1)(1−zx2)…(1⋯zxn+1)=S0+S1z+S2z2+…We have
S 1 = [ x 1 , x 2 , … , x n + 1 ; x i + n ] S 1 = x 1 , x 2 , … , x n + 1 ; x i + n S_(1)=[x_(1),x_(2),dots,x_(n+1);x^(i+n)]\mathrm{S}_{1}=\left[x_{1}, x_{2}, \ldots, x_{n+1} ; x^{i+n}\right]S1=[x1,x2,…,xn+1;xi+n]
If we write that the first member of (10) is equal to the product of the developments
1 1 − z x i = 1 + z x i + z 2 x i 2 + … + z m x i m + … , i = 1 , 2 , … , n + 1 1 1 − z x i = 1 + z x i + z 2 x i 2 + … + z m x i m + … , i = 1 , 2 , … , n + 1 (1)/(1-zx_(i))=1+zx_(i)+z^(2)x_(i)^(2)+dots+z^(m)x_(i)^(m)+dots,i=1,2,dots,n+1\frac{1}{1-z x_{i}}=1+z x_{i}+z^{2} x_{i}^{2}+\ldots+z^{m} x_{i}^{m}+\ldots, i=1,2, \ldots, n+111−zxi=1+zxi+z2xi2+…+zmxim+…,i=1,2,…,n+1
we find
S i = [ x 1 , x 2 , … , x n + 1 ; x i + n ] = Σ x 1 α 1 x 2 α 2 … x n + 1 α n + 1 S i = x 1 , x 2 , … , x n + 1 ; x i + n = Σ x 1 α 1 x 2 α 2 … x n + 1 α n + 1 S_(i)=[x_(1),x_(2),dots,x_(n+1);x^(i+n)]=Sigmax_(1)^(alpha_(1))x_(2)^(alpha_(2))dotsx_(n+1)^(alpha_(n+1))\mathrm{S}_{i}=\left[x_{1}, x_{2}, \ldots, x_{n+1} ; x^{i+n}\right]=\Sigma x_{1}^{\alpha_{1}} x_{2}^{\alpha_{2}} \ldots x_{n+1}^{\alpha_{n+1}}Si=[x1,x2,…,xn+1;xi+n]=Σx1α1x2α2…xn+1αn+1
the summation being extended to all the whole and positive solutions of the equation α 1 + α 2 + … + α n + 1 = i α 1 + α 2 + … + α n + 1 = i alpha_(1)+alpha_(2)+dots+alpha_(n+1)=i\alpha_{1}+\alpha_{2}+\ldots+\alpha_{n+1}=iα1+α2+…+αn+1=i.
3. The fundamental formula for transforming divided differences. Now consider the function f f fffset to m m mmmdistinct points
(11) x 1 , x 2 , … , x m ( m ≧ n + 1 ) (11) x 1 , x 2 , … , x m ( m ≧ n + 1 ) {:(11)x_(1)","x_(2)","dots","x_(m)quad(m >= n+1):}\begin{equation*} x_{1}, x_{2}, \ldots, x_{m} \quad(m \geqq n+1) \tag{11} \end{equation*}(11)x1,x2,…,xm(m≧n+1)
To simplify the notations we will set
(12) Δ j i ( f ) = [ x i , x i + 1 , … , x i + j ; f ] , Δ 0 i ( f ) = f ( x i ) , (13) ξ i , j + 1 ( x ) = ( x − x i ) ( x − x i + 1 ) … ( x − x i + j ) , ξ i , 0 ( x ) = 1 , i = 1 , 2 , … , m − j , j = 0 , 1 , … , m − 1 . (12) Δ j i ( f ) = x i , x i + 1 , … , x i + j ; f , Δ 0 i ( f ) = f x i , (13) ξ i , j + 1 ( x ) = x − x i x − x i + 1 … x − x i + j , ξ i , 0 ( x ) = 1 , i = 1 , 2 , … , m − j , j = 0 , 1 , … , m − 1 . {:[(12)Delta_(j)^(i)(f)=[x_(i),x_(i+1),dots,x_(i+j);f]","Delta_(0)^(i)(f)=f(x_(i))","],[(13)xi_(i,j+1)(x)=(x-x_(i))(x-x_(i+1))dots(x-x_(i+j))","quadxi_(i,0)(x)=1","],[i=1","2","dots","m-j","quad j=0","1","dots","m-1.]:}\begin{align*} \Delta_{j}^{i}(f) & =\left[x_{i}, x_{i+1}, \ldots, x_{i+j} ; f\right], \Delta_{0}^{i}(f)=f\left(x_{i}\right), \tag{12}\\ \xi_{i, j+1}(x) & =\left(x-x_{i}\right)\left(x-x_{i+1}\right) \ldots\left(x-x_{i+j}\right), \quad \xi_{i, 0}(x)=1, \tag{13}\\ i & =1,2, \ldots, m-j, \quad j=0,1, \ldots, m-1 . \end{align*}(12)ΔIi(f)=[xi,xi+1,…,xi+I;f],Δ0i(f)=f(xi),(13)ξi,I+1(x)=(x−xi)(x−xi+1)…(x−xi+I),ξi,0(x)=1,i=1,2,…,m−I,I=0,1,…,m−1.
So 1 , n + 1 ( x ) 1 , n + 1 ( x ) _(1,n+1)(x){ }_{1, n+1}(x)1,n+1(x)is precisely the polynomial (2) already considered. With these notations we have
Δ j i ( f ) = ∑ r = i i ⊬ j f ( x r ) φ i , j + 1 ′ ( x r ) Δ j i ( f ) = ∑ r = i i ⊬ j   f x r φ i , j + 1 ′ x r Delta_(j)^(i)(f)=sum_(r=i)^(i⊬j)(f(x_(r)))/(varphi_(i,j+1)^(')(x_(r)))\Delta_{j}^{i}(f)=\sum_{r=i}^{i \nvdash j} \frac{f\left(x_{r}\right)}{\varphi_{i, j+1}^{\prime}\left(x_{r}\right)}ΔIi(f)=∑r=ii⊬If(xr)φi,I+1′(xr)
We will now consider a linear and homogeneous expression of f ( x i ) f x i f(x_(i))f\left(x_{i}\right)f(xi)of the form
F = ∑ i = 1 m λ i f ( x i ) F = ∑ i = 1 m   λ i f x i F=sum_(i=1)^(m)lambda_(i)f(x_(i))\mathrm{F}=\sum_{i=1}^{m} \lambda_{i} f\left(x_{i}\right)F=∑i=1mλif(xi)
THE λ i λ i lambda_(i)\lambda_{i}λibeing independent of the function f f fff.
We can always put such an expression in the form
F = ∑ i = 1 n μ i Δ i − 1 1 ( f ) + ∑ i = 1 m − n v i Δ n i ( f ) F = ∑ i = 1 n   μ i Δ i − 1 1 ( f ) + ∑ i = 1 m − n   v i Δ n i ( f ) F=sum_(i=1)^(n)mu_(i)Delta_(i-1)^(1)(f)+sum_(i=1)^(m-n)v_(i)Delta_(n)^(i)(f)\mathrm{F}=\sum_{i=1}^{n} \mu_{i} \Delta_{i-1}^{1}(f)+\sum_{i=1}^{m-n} v_{i} \Delta_{n}^{i}(f)F=∑i=1nμiΔi−11(f)+∑i=1m−nviΔni(f)
where the μ l μ l mu_(l)\mu_{l}μLand the ν i ν i nu_(i)\nu_{i}νido not depend on the function t t ttt. These coefficients are completely determined by the coefficients λ r λ r lambda_(r)\lambda_{r}λr. Indeed, identification leads us to a linear system of m m mmmequations with respect to m m mmmunknowns μ l , ν l μ l , ν l mu_(l),nu_(l)\mu_{l}, \nu_{l}μL,νL. The determinant of this system
1 φ 1 , 2 ′ ( x 2 ) φ 1 , 3 ′ ( x 3 ) … φ 1 , n ′ ( x n ) φ 1 , n + 1 ′ ( x n + 1 ) φ 2 , n + 1 ′ ( x n + 2 ) … φ m − n , n + 1 ′ ( x m ) 1 φ 1 , 2 ′ x 2 φ 1 , 3 ′ x 3 … φ 1 , n ′ x n φ 1 , n + 1 ′ x n + 1 φ 2 , n + 1 ′ x n + 2 … φ m − n , n + 1 ′ x m (1)/(varphi_(1,2)^(')(x_(2))varphi_(1,3)^(')(x_(3))dotsvarphi_(1,n)^(')(x_(n))varphi_(1,n+1)^(')(x_(n+1))varphi_(2,n+1)^(')(x_(n+2))dotsvarphi_(m-n,n+1)^(')(x_(m)))\frac{1}{\varphi_{1,2}^{\prime}\left(x_{2}\right) \varphi_{1,3}^{\prime}\left(x_{3}\right) \ldots \varphi_{1, n}^{\prime}\left(x_{n}\right) \varphi_{1, n+1}^{\prime}\left(x_{n+1}\right) \varphi_{2, n+1}^{\prime}\left(x_{n+2}\right) \ldots \varphi_{m-n, n+1}^{\prime}\left(x_{m}\right)}1φ1,2′(x2)φ1,3′(x3)…φ1,n′(xn)φ1,n+1′(xn+1)φ2,n+1′(xn+2)…φm−n,n+1′(xm)is different from zero.
It remains to be seen how the coefficients are determined. μ i , y i μ i , y i mu_(i),y_(i)\mu_{i}, y_{i}μi,yiWe will obtain them by choosing the function appropriately. f f fff.
To have the coefficients μ i μ i mu_(i)\mu_{i}μi, let's take for f f fffthe polynomial φ 1 , j − 1 = φ 1 , j − 1 ( x ) φ 1 , j − 1 = φ 1 , j − 1 ( x ) varphi_(1,j-1)=varphi_(1,j-1)(x)\varphi_{1, j-1}=\varphi_{1, j-1}(x)φ1,I−1=φ1,I−1(x). We then have
Δ i − 1 1 ( φ 1 , j − 1 ) = { 0 , i = 1 , 2 , … , j − 1 1 , i = j 0 , i = j + 1 , j + 2 , … , n Δ n t ( ℑ 1 , j − 1 ) = 0 , i = 1 , 2 , … , m − n Δ i − 1 1 φ 1 , j − 1 = 0 , i = 1 , 2 , … , j − 1 1 , i = j 0 , i = j + 1 , j + 2 , … , n Δ n t ℑ 1 , j − 1 = 0 , i = 1 , 2 , … , m − n {:[Delta_(i-1)^(1)(varphi_(1,j-1))={[0",",i=1","2","dots","j-1],[1",",i=j],[0",",i=j+1","j+2","dots","n]:}],[Delta_(n)^(t)(ℑ_(1,j-1))=0","i=1","2","dots","m-n]:}\begin{aligned} \Delta_{i-1}^{1}\left(\varphi_{1, j-1}\right) & = \begin{cases}0, & i=1,2, \ldots, j-1 \\ 1, & i=j \\ 0, & i=j+1, j+2, \ldots, n\end{cases} \\ \Delta_{n}^{t}\left(\Im_{1, j-1}\right) & =0, i=1,2, \ldots, m-n \end{aligned}Δi−11(φ1,I−1)={0,i=1,2,…,I−11,i=I0,i=I+1,I+2,…,nΔnt(No.1,I−1)=0,i=1,2,…,m−n
SO
(14) μ j = ∑ i = 1 m λ i f 1 , j − 1 ( x l ) = ∑ i = j m λ i ( x i − x 1 ) ( x i − x 2 ) … ( x i − x j − 1 ) j = 1 , 2 , … , n . (14) μ j = ∑ i = 1 m   λ i f 1 , j − 1 x l = ∑ i = j m   λ i x i − x 1 x i − x 2 … x i − x j − 1 j = 1 , 2 , … , n . {:[(14)mu_(j)=sum_(i=1)^(m)lambda_(i)f_(1,j-1)(x_(l))=sum_(i=j)^(m)lambda_(i)(x_(i)-x_(1))(x_(i)-x_(2))dots(x_(i)-x_(j-1))],[j=1","2","dots","n.]:}\begin{gather*} \mu_{j}=\sum_{i=1}^{m} \lambda_{i} f_{1, j-1}\left(x_{l}\right)=\sum_{i=j}^{m} \lambda_{i}\left(x_{i}-x_{1}\right)\left(x_{i}-x_{2}\right) \ldots\left(x_{i}-x_{j-1}\right) \tag{14}\\ j=1,2, \ldots, n . \end{gather*}(14)μI=∑i=1mλif1,I−1(xL)=∑i=Imλi(xi−x1)(xi−x2)…(xi−xI−1)I=1,2,…,n.
To obtain the coefficients % i % i %_(i)\%_{i}%i, we take for f f fffthe following functions
(15) f j ∗ = f j ∗ ( x ) = { 0 , pour x = x 1 , x 2 , … , x j + n − 1 φ j + 1 , n − 1 ( x ) , pour x = x j + n , x j + n + 1 , … , x m j = 1 , 2 , … , m − n (15) f j ∗ = f j ∗ ( x ) = 0 ,  pour  x = x 1 , x 2 , … , x j + n − 1 φ j + 1 , n − 1 ( x ) ,  pour  x = x j + n , x j + n + 1 , … , x m j = 1 , 2 , … , m − n {:(15)f_(j)^(**)=f_(j)^(**)(x)={[0","," pour "x=x_(1)","x_(2)","dots","x_(j+n-1)],[varphi_(j+1,n-1)(x)","," pour "x=x_(j+n)","x_(j+n+1)","dots","x_(m)],[j=1","2","dots",",m-n]:}:}\begin{align*} f_{j}^{*}=f_{j}^{*}(x)= & \begin{cases}0, & \text { pour } x=x_{1}, x_{2}, \ldots, x_{j+n-1} \\ \varphi_{j+1, n-1}(x), & \text { pour } x=x_{j+n}, x_{j+n+1}, \ldots, x_{m} \\ j=1,2, \ldots, & m-n\end{cases} \tag{15} \end{align*}(15)fI∗=fI∗(x)={0, For x=x1,x2,…,xI+n−1φI+1,n−1(x), For x=xI+n,xI+n+1,…,xmI=1,2,…,m−n
We have
Δ i − 1 1 ( f j ∗ ) = 0 , i = 1 , 2 , … , n Δ n i ( f j ∗ ) = { 0 , i = 1 , 2 , … , j − 1 Δ n j ( f j ∗ ) , i = j 0 , i = j + 1 , j + 2 , … , m − n Δ i − 1 1 f j ∗ = 0 , i = 1 , 2 , … , n Δ n i f j ∗ = 0 , i = 1 , 2 , … , j − 1 Δ n j f j ∗ , i = j 0 , i = j + 1 , j + 2 , … , m − n {:[Delta_(i-1)^(1)(f_(j)^(**))=0","i=1","2","dots","n],[Delta_(n)^(i)(f_(j)^(**))={[0",",i=1","2","dots","j-1],[Delta_(n)^(j)(f_(j)^(**))",",i=j],[0",",i=j+1","j+2","dots","m-n]:}]:}\begin{aligned} \Delta_{i-1}^{1}\left(f_{j}^{*}\right) & =0, i=1,2, \ldots, n \\ \Delta_{n}^{i}\left(f_{j}^{*}\right) & = \begin{cases}0, & i=1,2, \ldots, j-1 \\ \Delta_{n}^{j}\left(f_{j}^{*}\right), & i=j \\ 0, & i=j+1, j+2, \ldots, m-n\end{cases} \end{aligned}Δi−11(fI∗)=0,i=1,2,…,nΔni(fI∗)={0,i=1,2,…,I−1ΔnI(fI∗),i=I0,i=I+1,I+2,…,m−n
SO
F = y j Δ n j ( f j ∗ ) = f j ∗ ( x j + n ) ( x j + n − x j ) ( x j + n − x j + 1 ) … ( x j + n − x j + n − 1 ) = v j x j + n − x j F = y j Δ n j f j ∗ = f j ∗ x j + n x j + n − x j x j + n − x j + 1 … x j + n − x j + n − 1 = v j x j + n − x j F=y_(j)Delta_(n)^(j)(f_(j)^(**))=(f_(j)^(**)(x_(j+n)))/((x_(j+n)-x_(j))(x_(j+n)-x_(j+1))dots(x_(j+n)-x_(j+n-1)))=(v_(j))/(x_(j+n)-x_(j))\mathrm{F}=y_{j} \Delta_{n}^{j}\left(f_{j}^{*}\right)=\frac{f_{j}^{*}\left(x_{j+n}\right)}{\left(x_{j+n}-x_{j}\right)\left(x_{j+n}-x_{j+1}\right) \ldots\left(x_{j+n}-x_{j+n-1}\right)}=\frac{v_{j}}{x_{j+n}-x_{j}}F=yIΔnI(fI∗)=fI∗(xI+n)(xI+n−xI)(xI+n−xI+1)…(xI+n−xI+n−1)=vIxI+n−xI.
We deduce from this
(16) y j = ( x j + n − x j ) ∑ i = 1 m λ i f j ∗ ( x i ) = ( x j + n − x j ) ∑ i = j + n m λ i φ j + 1 , n − 1 ( x i ) = = ( x j + n − x j ) ∑ i = j + n m λ i ( x i − x j + 1 ) ( x i − x j + 2 ) … ( x i − x j + n − 1 ) . (16) y j = x j + n − x j ∑ i = 1 m   λ i f j ∗ x i = x j + n − x j ∑ i = j + n m   λ i φ j + 1 , n − 1 x i = = x j + n − x j ∑ i = j + n m   λ i x i − x j + 1 x i − x j + 2 … x i − x j + n − 1 . {:[(16)y_(j)=(x_(j+n)-x_(j))sum_(i=1)^(m)lambda_(i)f_(j)^(**)(x_(i))=(x_(j+n)-x_(j))sum_(i=j+n)^(m)lambda_(i)varphi_(j+1,n-1)(x_(i))=],[=(x_(j+n)-x_(j))sum_(i=j+n)^(m)lambda_(i)(x_(i)-x_(j+1))(x_(i)-x_(j+2))dots(x_(i)-x_(j+n-1)).]:}\begin{align*} y_{j} & =\left(x_{j+n}-x_{j}\right) \sum_{i=1}^{m} \lambda_{i} f_{j}^{*}\left(x_{i}\right)=\left(x_{j+n}-x_{j}\right) \sum_{i=j+n}^{m} \lambda_{i} \varphi_{j+1, n-1}\left(x_{i}\right)= \tag{16}\\ & =\left(x_{j+n}-x_{j}\right) \sum_{i=j+n}^{m} \lambda_{i}\left(x_{i}-x_{j+1}\right)\left(x_{i}-x_{j+2}\right) \ldots\left(x_{i}-x_{j+n-1}\right) . \end{align*}(16)yI=(xI+n−xI)∑i=1mλifI∗(xi)=(xI+n−xI)∑i=I+nmλiφI+1,n−1(xi)==(xI+n−xI)∑i=I+nmλi(xi−xI+1)(xi−xI+2)…(xi−xI+n−1).
The fundamental transformation formula is therefore finally written in the following form
(17) ∑ i = 1 m λ i f ( x i ) = ∑ j = 1 n [ ∑ i = j m λ i ρ 1 , j − 1 ( x i ) ] Δ j − 1 1 ( f ) + + ∑ j = 1 m − n [ ( x j + n − x j ) ∑ i = j + n m λ i φ j + 1 , n − 1 ( x l ) ] Δ n j ( f ) (17) ∑ i = 1 m   λ i f x i = ∑ j = 1 n   ∑ i = j m   λ i ρ 1 , j − 1 x i Δ j − 1 1 ( f ) + + ∑ j = 1 m − n   x j + n − x j ∑ i = j + n m   λ i φ j + 1 , n − 1 x l Δ n j ( f ) {:[(17)sum_(i=1)^(m)lambda_(i)f(x_(i))=sum_(j=1)^(n)[sum_(i=j)^(m)lambda_(i)rho_(1,j-1)(x_(i))]Delta_(j-1)^(1)(f)+],[+sum_(j=1)^(m-n)[(x_(j+n)-x_(j))sum_(i=j+n)^(m)lambda_(i)varphi_(j+1,n-1)(x_(l))]Delta_(n)^(j)(f)]:}\begin{align*} & \sum_{i=1}^{m} \lambda_{i} f\left(x_{i}\right)=\sum_{j=1}^{n}\left[\sum_{i=j}^{m} \lambda_{i} \rho_{1, j-1}\left(x_{i}\right)\right] \Delta_{j-1}^{1}(f)+ \tag{17}\\ & +\sum_{j=1}^{m-n}\left[\left(x_{j+n}-x_{j}\right) \sum_{i=j+n}^{m} \lambda_{i} \varphi_{j+1, n-1}\left(x_{l}\right)\right] \Delta_{n}^{j}(f) \end{align*}(17)∑i=1mλif(xi)=∑I=1n[∑i=Imλiρ1,I−1(xi)]ΔI−11(f)++∑I=1m−n[(xI+n−xI)∑i=I+nmλiφI+1,n−1(xL)]ΔnI(f)
For any natural number n n nnnwe have such a formula. The coefficients μ i , ν i μ i , ν i mu_(i),nu_(i)\mu_{i}, \nu_{i}μi,νican also be obtained by recurrence relations. If we denote by μ i ( n ) , v i ( n ) μ i ( n ) , v i ( n ) mu_(i)^((n)),v_(i)^((n))\mu_{i}^{(n)}, v_{i}^{(n)}μi(n),vi(n)these coefficients, to highlight the number n n nnn, We have
μ j ( n ) = μ j ( n − 1 ) , j = 1 , 2 , … , n − 1 μ n ( n ) = v 1 ( n − 1 ) + v 2 ( n − 1 ) + … … + v m − n + 1 ( n − 1 ) v j ( n ) = ( x j + n − x j ) [ v j + 1 ( n − 1 ) + v j + 2 ( n − 1 ) + … + v m − n + 1 ( n − 1 ) ] j = 1 , 2 , … , m − n μ j ( n ) = μ j ( n − 1 ) , j = 1 , 2 , … , n − 1 μ n ( n ) = v 1 ( n − 1 ) + v 2 ( n − 1 ) + … … + v m − n + 1 ( n − 1 ) v j ( n ) = x j + n − x j v j + 1 ( n − 1 ) + v j + 2 ( n − 1 ) + … + v m − n + 1 ( n − 1 ) j = 1 , 2 , … , m − n {:[mu_(j)^((n))=mu_(j)^((n-1))","j=1","2","dots","n-1],[mu_(n)^((n))=v_(1)^((n-1))+v_(2)^((n-1))+dots dots+v_(m-n+1)^((n-1))],[v_(j)^((n))=(x_(j+n)-x_(j))[v_(j+1)^((n-1))+v_(j+2)^((n-1))+dots+v_(m-n+1)^((n-1))]],[j=1","2","dots","m-n]:}\begin{gathered} \mu_{j}^{(n)}=\mu_{j}^{(n-1)}, j=1,2, \ldots, n-1 \\ \mu_{n}^{(n)}=v_{1}^{(n-1)}+v_{2}^{(n-1)}+\ldots \ldots+v_{m-n+1}^{(n-1)} \\ v_{j}^{(n)}=\left(x_{j+n}-x_{j}\right)\left[v_{j+1}^{(n-1)}+v_{j+2}^{(n-1)}+\ldots+v_{m-n+1}^{(n-1)}\right] \\ j=1,2, \ldots, m-n \end{gathered}μI(n)=μI(n−1),I=1,2,…,n−1μn(n)=v1(n−1)+v2(n−1)+……+vm−n+1(n−1)vI(n)=(xI+n−xI)[vI+1(n−1)+vI+2(n−1)+…+vm−n+1(n−1)]I=1,2,…,m−n

4.--Some applications of the fundamental formula.

I. Let f = f ( x ) , g = g ( x ) f = f ( x ) , g = g ( x ) f=f(x),g=g(x)f=f(x), g=g(x)f=f(x),g=g(x)two functions defined on the points x i x i x_(i)x_{i}xi. Consider the divided difference
F = [ x 1 , x 2 , … , x n + 1 ; f g ] F = x 1 , x 2 , … , x n + 1 ; f g F=[x_(1),x_(2),dots,x_(n+1);fg]\mathrm{F}=\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f g\right]F=[x1,x2,…,xn+1;fg]
of the product f g f g fgf gfg. In this case we have
λ i = g ( x i ) g 1 n + 1 ′ ( x i ) , i = 1 , 2 , … , n + 1 , λ i = 0 , i > n + 1 . λ i = g x i g 1 n + 1 ′ x i , i = 1 , 2 , … , n + 1 , λ i = 0 , i > n + 1 . lambda_(i)=(g(x_(i)))/(g_(1n+1)^(')(x_(i))),i=1,2,dots,n+1,lambda_(i)=0,i > n+1.\lambda_{i}=\frac{g\left(x_{i}\right)}{g_{1 n+1}^{\prime}\left(x_{i}\right)}, i=1,2, \ldots, n+1, \lambda_{i}=0, i>n+1 .λi=g(xi)g1n+1′(xi),i=1,2,…,n+1,λi=0,i>n+1.
If we notice that ′ 1 , n + 1 = φ 1 , j − 1 ⋅ φ j , n − j + 2 ′ 1 , n + 1 = φ 1 , j − 1 ⋅ φ j , n − j + 2 ^(')_(1,n+1)=varphi_(1,j-1)*varphi_(j,n-j+2){ }^{\prime}{ }_{1, n+1}=\varphi_{1, j-1} \cdot \varphi_{j, n-j+2}′1,n+1=φ1,I−1⋅φI,n−I+2, we deduce from this
φ 1 , j − 1 ( x i ) = φ 1 , n + 1 ′ ( x i ) φ j , n − j + 2 ′ ( x i ) , pour j ≦ i ≦ n + 1 φ 1 , j − 1 x i = φ 1 , n + 1 ′ x i φ j , n − j + 2 ′ x i , pour  j ≦ i ≦ n + 1 varphi_(1,j-1)(x_(i))=(varphi_(1,n+1)^(')(x_(i)))/(varphi_(j,n-j+2)^(')(x_(i)))", pour "j <= i <= n+1\varphi_{1, j-1}\left(x_{i}\right)=\frac{\varphi_{1, n+1}^{\prime}\left(x_{i}\right)}{\varphi_{j, n-j+2}^{\prime}\left(x_{i}\right)} \text {, pour } j \leqq i \leqq n+1φ1,I−1(xi)=φ1,n+1′(xi)φI,n−I+2′(xi), For I≦i≦n+1
and formula (14) gives us
μ j = ∑ i = j n + 1 g ( x i ) ψ j , n − j + 2 ( x i ) = [ x j , x j + 1 , … , x n + 1 ; g ] , j = 1 , 2 , … , n . μ j = ∑ i = j n + 1   g x i ψ j , n − j + 2 x i = x j , x j + 1 , … , x n + 1 ; g , j = 1 , 2 , … , n . mu_(j)=sum_(i=j)^(n+1)(g(x_(i)))/(psi_(j,n-j+2)(x_(i)))=[x_(j),x_(j+1),dots,x_(n+1);g],j=1,2,dots,n.\mu_{j}=\sum_{i=j}^{n+1} \frac{g\left(x_{i}\right)}{\psi_{j, n-j+2}\left(x_{i}\right)}=\left[x_{j}, x_{j+1}, \ldots, x_{n+1} ; g\right], j=1,2, \ldots, n .μI=∑i=In+1g(xi)ψI,n−I+2(xi)=[xI,xI+1,…,xn+1;g],I=1,2,…,n.
Formula (16) gives us
v 1 = g ( x n + 1 ) , v j = 0 , j = 2 , 3 , … v 1 = g x n + 1 , v j = 0 , j = 2 , 3 , … v_(1)=g(x_(n+1)),v_(j)=0,j=2,3,dotsv_{1}=g\left(x_{n+1}\right), v_{j}=0, j=2,3, \ldotsv1=g(xn+1),vI=0,I=2,3,…
As a result, we have the following formula
(18) [ x 1 , x 2 , … , x n + 1 ; f g ] = ∑ i = 1 n + 1 [ x 1 , x 2 , … , x i ; f ] ⌊ x i , x i + 1 , … x n + 1 ; g ] x 1 , x 2 , … , x n + 1 ; f g = ∑ i = 1 n + 1   x 1 , x 2 , … , x i ; f x i , x i + 1 , … x n + 1 ; g [x_(1),x_(2),dots,x_(n+1);fg]=sum_(i=1)^(n+1)[x_(1),x_(2),dots,x_(i);f]|__x_(i),x_(i+1),dotsx_(n+1);g]\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f g\right]=\sum_{i=1}^{n+1}\left[x_{1}, x_{2}, \ldots, x_{i} ; f\right]\left\lfloor x_{i}, x_{i+1}, \ldots x_{n+1} ; g\right][x1,x2,…,xn+1;fg]=∑i=1n+1[x1,x2,…,xi;f]⌊xi,xi+1,…xn+1;g], which gives the divided difference of a product. This is the generalization of Leibniz's formula.
II. The Lagrange polynomial (1) is of the form F. A simple calculation shows us that in this case
λ i = S 1 , n + 1 ( x ) ( x − x i ) P 1 , n + 1 ′ ( x i ) , i = 1 , 2 , … , n + 1 , λ i = 0 , i > n + 1 . λ i = S 1 , n + 1 ( x ) x − x i P 1 , n + 1 ′ x i , i = 1 , 2 , … , n + 1 , λ i = 0 , i > n + 1 . lambda_(i)=(S_(1,n+1)(x))/((x-x_(i))P_(1,n+1)^(')(x_(i))),i=1,2,dots,n+1,lambda_(i)=0,i > n+1.\lambda_{i}=\frac{\mathrm{S}_{1, n+1}(x)}{\left(x-x_{i}\right) \mathrm{P}_{1, n+1}^{\prime}\left(x_{i}\right)}, i=1,2, \ldots, n+1, \lambda_{i}=0, i>n+1 .λi=S1,n+1(x)(x−xi)P1,n+1′(xi),i=1,2,…,n+1,λi=0,i>n+1.
Formula (14) gives us
μ j = P ( x 1 , x 2 , … , x n + 1 ; φ 1 , j − 1 ∣ x ) = φ 1 , j − 1 ( x ) , j = 1 , 2 , … , n μ j = P x 1 , x 2 , … , x n + 1 ; φ 1 , j − 1 ∣ x = φ 1 , j − 1 ( x ) , j = 1 , 2 , … , n mu_(j)=P(x_(1),x_(2),dots,x_(n+1);varphi_(1,j-1)∣x)=varphi_(1,j-1)(x),j=1,2,dots,n\mu_{j}=\mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; \varphi_{1, j-1} \mid x\right)=\varphi_{1, j-1}(x), \mathrm{j}=1,2, \ldots, nμI=P(x1,x2,…,xn+1;φ1,I−1∣x)=φ1,I−1(x),I=1,2,…,n
And
y 1 = ( x n + 1 − x 1 ) P ( x 1 , x 2 , … , x n + 1 ; f 1 ∗ ∣ x ) = = ( x n + 1 − x 1 ) ∗ 1 , n ( x ) f 1 ∗ ( x n + 1 ) ξ 1 , n + 1 ′ ( x n + 1 ) = ξ 1 , n ( x ) , y i = 0 , i > 1 y 1 = x n + 1 − x 1 P x 1 , x 2 , … , x n + 1 ; f 1 ∗ ∣ x = = x n + 1 − x 1 ∗ 1 , n ( x ) f 1 ∗ x n + 1 ξ 1 , n + 1 ′ x n + 1 = ξ 1 , n ( x ) , y i = 0 , i > 1 {:[y_(1)=(x_(n+1)-x_(1))P(x_(1),x_(2),dots,x_(n+1);f_(1)^(**)∣x)=],[=(x_(n+1)-x_(1))_(**_(1,n))(x)(f_(1)^(**)(x_(n+1)))/(xi_(1,n+1)^(')(x_(n+1)))=xi_(1,n)(x)","y_(i)=0","i > 1]:}\begin{gathered} y_{1}=\left(x_{n+1}-x_{1}\right) \mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f_{1}^{*} \mid x\right)= \\ =\left(x_{n+1}-x_{1}\right){ }_{*_{1, n}}(x) \frac{f_{1}^{*}\left(x_{n+1}\right)}{\xi_{1, n+1}^{\prime}\left(x_{n+1}\right)}=\xi_{1, n}(x), y_{i}=0, i>1 \end{gathered}y1=(xn+1−x1)P(x1,x2,…,xn+1;f1∗∣x)==(xn+1−x1)∗1,n(x)f1∗(xn+1)ξ1,n+1′(xn+1)=ξ1,n(x),yi=0,i>1
and we get the well-known interpolation formula
P ( x 1 , x 2 , … , x n + 1 ; f ∣ x ) = f ( x 1 ) + ( x − x 1 ) [ x 1 , x 2 ; f ] + + ( x − x 1 ) ( x − x 2 ) [ x 1 , x 2 , x 3 ; f ] + … + ( x − x 1 ) ( x − x 2 ) … ( x − x n ) [ x 1 , x 2 , … , x n + 1 ; f ] P x 1 , x 2 , … , x n + 1 ; f ∣ x = f x 1 + x − x 1 x 1 , x 2 ; f + + x − x 1 x − x 2 x 1 , x 2 , x 3 ; f + … + x − x 1 x − x 2 … x − x n x 1 , x 2 , … , x n + 1 ; f {:[P(x_(1),x_(2),dots,x_(n+1);f∣x)=f(x_(1))+(x-x_(1))[x_(1),x_(2);f]+],[+(x-x_(1))(x-x_(2))[x_(1),x_(2),x_(3);f]+dots],[+(x-x_(1))(x-x_(2))dots(x-x_(n))[x_(1),x_(2),dots,x_(n+1);f]]:}\begin{gathered} \mathrm{P}\left(x_{1}, x_{2}, \ldots, x_{n+1} ; f \mid x\right)=f\left(x_{1}\right)+\left(x-x_{1}\right)\left[x_{1}, x_{2} ; f\right]+ \\ +\left(x-x_{1}\right)\left(x-x_{2}\right)\left[x_{1}, x_{2}, x_{3} ; f\right]+\ldots \\ +\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{n}\right)\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right] \end{gathered}P(x1,x2,…,xn+1;f∣x)=f(x1)+(x−x1)[x1,x2;f]++(x−x1)(x−x2)[x1,x2,x3;f]+…+(x−x1)(x−x2)…(x−xn)[x1,x2,…,xn+1;f]
III. Let us take
F = [ x n + 1 , x n + 2 , … , x 2 n ; f ] − [ x 1 , x 2 , … , x n ; f ] . F = x n + 1 , x n + 2 , … , x 2 n ; f − x 1 , x 2 , … , x n ; f . F=[x_(n+1),x_(n+2),dots,x_(2n);f]-[x_(1),x_(2),dots,x_(n);f].\mathrm{F}=\left[x_{n+1}, x_{n+2}, \ldots, x_{2 n} ; f\right]-\left[x_{1}, x_{2}, \ldots, x_{n} ; f\right] .F=[xn+1,xn+2,…,x2n;f]−[x1,x2,…,xn;f].
It is easily verified that μ j = 0 , j = 1 , 2 , … , n μ j = 0 , j = 1 , 2 , … , n mu_(j)=0,j=1,2,dots,n\mu_{j}=0, j=1,2, \ldots, nμI=0,I=1,2,…,n. We find
y j = ( x j + n − x j ) { [ x n + 1 , x n + 2 , … , x 2 n ; f j ∗ ] − [ x 1 , x 2 , . , x n ; f j ∗ ] } = = ( x j + n − x n ) [ x n + 1 , x n + 2 , … , x 2 n ; P j + 1 , n − 1 ] = ( x j + n − x j ) y j = x j + n − x j x n + 1 , x n + 2 , … , x 2 n ; f j ∗ − x 1 , x 2 , . , x n ; f j ∗ = = x j + n − x n x n + 1 , x n + 2 , … , x 2 n ; P j + 1 , n − 1 = x j + n − x j {:[y_(j)=(x_(j+n)-x_(j)){[x_(n+1),x_(n+2),dots,x_(2n);f_(j)^(**)]-[x_(1),x_(2),.,x_(n);f_(j)^(**)]}=],[=(x_(j+n)-x_(n))[x_(n+1),x_(n+2),dots,x_(2n);P_(j+1,n-1)]=(x_(j+n)-x_(j))]:}\begin{aligned} y_{j}= & \left(x_{j+n}-x_{j}\right)\left\{\left[x_{n+1}, x_{n+2}, \ldots, x_{2 n} ; f_{j}^{*}\right]-\left[x_{1}, x_{2}, ., x_{n} ; f_{j}^{*}\right]\right\}= \\ = & \left(x_{j+n}-x_{n}\right)\left[x_{n+1}, x_{n+2}, \ldots, x_{2 n} ; \mathscr{P}_{j+1, n-1}\right]=\left(x_{j+n}-x_{j}\right) \end{aligned}yI=(xI+n−xI){[xn+1,xn+2,…,x2n;fI∗]−[x1,x2,.,xn;fI∗]}==(xI+n−xn)[xn+1,xn+2,…,x2n;PI+1,n−1]=(xI+n−xI)
and we have, with the notations (12), the following formula
(19) Δ n − 1 n + 1 ( t ) − Δ n − 1 1 ( f ) = ∑ j = 1 n ( x j + n − x j ) Δ n j ( f ) (19) Δ n − 1 n + 1 ( t ) − Δ n − 1 1 ( f ) = ∑ j = 1 n   x j + n − x j Δ n j ( f ) {:(19)Delta_(n-1)^(n+1)(t)-Delta_(n-1)^(1)(f)=sum_(j=1)^(n)(x_(j+n)-x_(j))Delta_(n)^(j)(f):}\begin{equation*} \Delta_{n-1}^{n+1}(t)-\Delta_{n-1}^{1}(f)=\sum_{j=1}^{n}\left(x_{j+n}-x_{j}\right) \Delta_{n}^{j}(f) \tag{19} \end{equation*}(19)Δn−1n+1(t)−Δn−11(f)=∑I=1n(xI+n−xI)ΔnI(f)
which can, moreover, be established just as simply using the recurrence formula (6)
IV. Let
x i 1 , x i 2 , … , x i n + 1 ( 1 ≦ i 1 < i 2 < … < i n + 1 ≦ m ) x i 1 , x i 2 , … , x i n + 1 1 ≦ i 1 < i 2 < … < i n + 1 ≦ m x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1))quad(1 <= i_(1) < i_(2) < dots < i_(n+1) <= m)x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} \quad\left(1 \leqq i_{1}<i_{2}<\ldots<i_{n+1} \leqq m\right)xi1,xi2,…,xin+1(1≦i1<i2<…<in+1≦m)
a partial sequence extracted from the sequence (11). Let us take
F = [ x i 1 , x i 2 , … , x i n + 1 ; f ] F = x i 1 , x i 2 , … , x i n + 1 ; f F=[x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));f]\mathrm{F}=\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} ; f\right]F=[xi1,xi2,…,xin+1;f]
The fordamental formula (17) then gives us
(20) [ x i 1 , x i 2 , … , x i n + 1 ; f ] = ∑ j = 4 m − n ( x j + n − x j ) [ x i 1 , x i 2 , … , x i n + 1 ; f j ∗ ] Δ n j ( f ) x i 1 , x i 2 , … , x i n + 1 ; f = ∑ j = 4 m − n   x j + n − x j x i 1 , x i 2 , … , x i n + 1 ; f j ∗ Δ n j ( f ) [x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));f]=sum_(j=4)^(m-n)(x_(j+n)-x_(j))[x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));f_(j)^(**)]Delta_(n)^(j)(f)\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} ; f\right]=\sum_{j=4}^{m-n}\left(x_{j+n}-x_{j}\right)\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} ; f_{j}^{*}\right] \Delta_{n}^{j}(f)[xi1,xi2,…,xin+1;f]=∑I=4m−n(xI+n−xI)[xi1,xi2,…,xin+1;fI∗]ΔnI(f).
Remarks. 1 0 1 0 1^(0)1^{0}10. When F is identically zero for any polynomial of degree r − 1 r − 1 r-1r-1r−1We have μ j = 0 , j = 1 , 2 , … , r μ j = 0 , j = 1 , 2 , … , r mu_(j)=0,j=1,2,dots,r\mu_{j}=0, j=1,2, \ldots, rμI=0,I=1,2,…,r.
2 0 2 0 2^(0)2^{0}20. Formula (19) is valid even if the x i x i x_(i)x_{i}xiare not all distinct. It is sufficient that each of the sequences
x 1 , x 2 , … , x n ; x n + 1 , x n + 2 , … , x 2 n , x j , x j + 1 , … , x j + n , j = 1 , 2 , … , n , x 1 , x 2 , … , x n ; x n + 1 , x n + 2 , … , x 2 n , x j , x j + 1 , … , x j + n , j = 1 , 2 , … , n , {:[x_(1)","x_(2)","dots","x_(n);quadx_(n+1)","x_(n+2)","dots","x_(2n)","],[x_(j)","x_(j+1)","dots","x_(j+n)","quad j=1","2","dots","n","]:}\begin{aligned} & x_{1}, x_{2}, \ldots, x_{n} ; \quad x_{n+1}, x_{n+2}, \ldots, x_{2 n}, \\ & x_{j}, x_{j+1}, \ldots, x_{j+n}, \quad j=1,2, \ldots, n, \end{aligned}x1,x2,…,xn;xn+1,xn+2,…,x2n,xI,xI+1,…,xI+n,I=1,2,…,n,
is formed by distinct points. We deduce that if
α 1 , α 2 , … , α j , β 1 , β 2 , … , β j , α j + 1 = β j + 1 , α j + 2 = β j + 2 , … , α n = β n α 1 , α 2 , … , α j , β 1 , β 2 , … , β j , α j + 1 = β j + 1 , α j + 2 = β j + 2 , … , α n = β n alpha_(1),alpha_(2),dots,alpha_(j),beta_(1),beta_(2),dots,beta_(j),alpha_(j+1)=beta_(j+1),alpha_(j+2)=beta_(j+2),dots,alpha_(n)=beta_(n)\alpha_{1}, \alpha_{2}, \ldots, \alpha_{j}, \beta_{1}, \beta_{2}, \ldots, \beta_{j}, \alpha_{j+1}=\beta_{j+1}, \alpha_{j+2}=\beta_{j+2}, \ldots, \alpha_{n}=\beta_{n}α1,α2,…,αI,β1,β2,…,βI,αI+1=βI+1,αI+2=βI+2,…,αn=βn
are distinct points, we have the following formula
[ α 1 , α 2 , … , α n ; f ] − [ β 1 , β 2 , … , β n ; f ] = = ∑ i = 1 j ( α i − β i ) [ α 1 , α 2 , … , α i , β i , β i + 1 , … , β n ; f ] α 1 , α 2 , … , α n ; f − β 1 , β 2 , … , β n ; f = = ∑ i = 1 j   α i − β i α 1 , α 2 , … , α i , β i , β i + 1 , … , β n ; f {:[[alpha_(1),alpha_(2),dots,alpha_(n);f]-[beta_(1),beta_(2),dots,beta_(n);f]=],[quad=sum_(i=1)^(j)(alpha_(i)-beta_(i))[alpha_(1),alpha_(2),dots,alpha_(i),beta_(i),beta_(i+1),dots,beta_(n);f]]:}\begin{aligned} & {\left[\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n} ; f\right]-\left[\beta_{1}, \beta_{2}, \ldots, \beta_{n} ; f\right]=} \\ & \quad=\sum_{i=1}^{j}\left(\alpha_{i}-\beta_{i}\right)\left[\alpha_{1}, \alpha_{2}, \ldots, \alpha_{i}, \beta_{i}, \beta_{i+1}, \ldots, \beta_{n} ; f\right] \end{aligned}[α1,α2,…,αn;f]−[β1,β2,…,βn;f]==∑i=1I(αi−βi)[α1,α2,…,αi,βi,βi+1,…,βn;f]
  1. The formula for the mean of the divided differences. We will now assume that the points (11) are ordered, so that
x 1 < x 2 < … < x m . x 1 < x 2 < … < x m . x_(1) < x_(2) < dots < x_(m).x_{1}<x_{2}<\ldots<x_{m} .x1<x2<…<xm.
The determinant V ( x 1 , x 2 , … , x n + 1 ) V x 1 , x 2 , … , x n + 1 V(x_(1),x_(2),dots,x_(n+1))\mathrm{V}\left(x_{1}, x_{2}, \ldots, x_{n+1}\right)V(x1,x2,…,xn+1)then has a positive value and if we set
sg α = { − 1 , α < 0 , 0 , α = 0 , 1 , α > 0 , sg α = − 1 , α < 0 , 0 , α = 0 , 1 , α > 0 , sg alpha={[-1","quad alpha < 0","],[0","quad alpha=0","],[1","quad alpha > 0","]:}\operatorname{sg} \alpha=\left\{\begin{array}{r} -1, \quad \alpha<0, \\ 0, \quad \alpha=0, \\ 1, \quad \alpha>0, \end{array}\right.sgα={−1,α<0,0,α=0,1,α>0,
We have
s g [ x 1 , x 2 , … , x n + 1 ; f ] = sgU ( x 1 , x 2 , … x n + 1 ; f ) s g x 1 , x 2 , … , x n + 1 ; f = sgU x 1 , x 2 , … x n + 1 ; f sg[x_(1),x_(2),dots,x_(n+1);f]=sgU(x_(1),x_(2),dotsx_(n+1);f)s g\left[x_{1}, x_{2}, \ldots, x_{n+1} ; f\right]=\operatorname{sgU}\left(x_{1}, x_{2}, \ldots x_{n+1} ; f\right)sg[x1,x2,…,xn+1;f]=sgU(x1,x2,…xn+1;f)
Let us then take up formula (20). The following x l 1 , x i 2 , … , x i n + 1 x l 1 , x i 2 , … , x i n + 1 x_(l_(1)),x_(i_(2)),dots,x_(i_(n+1))x_{l_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}}xL1,xi2,…,xin+1is also ordered and we have i n + 1 − i 1 = p ≧ n i n + 1 − i 1 = p ≧ n i_(n+1)-i_(1)=p >= ni_{n+1}-i_{1}=p \geqq nin+1−i1=p≧n. We will demonstrate that then all the coefficients of Δ n j ( f ) Δ n j ( f ) Delta_(n)^(j)(f)\Delta_{n}^{j}(f)ΔnI(f)are ≥ 0 ≥ 0 >= 0\geq 0≥0The demonstration can be done by induction on the number p p ppp. The property is obvious to p = n p = n p=np=np=n, because in this case [ x i 1 , x i 2 , … , x i n + 1 ; f ] = Δ n i 1 ( f ) x i 1 , x i 2 , … , x i n + 1 ; f = Δ n i 1 ( f ) [x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));f]=Delta_(n)^(i_(1))(f)\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} ; f\right]=\Delta_{n}^{i_{1}}(f)[xi1,xi2,…,xin+1;f]=Δni1(f).
For p = n + 1 p = n + 1 p=n+1p=n+1p=n+1, let's take
i 1 = i , i 2 = i + 1 , … , i j = i + j − 1 i j + 1 = i + j + 1 , i j + 2 = i + j + 2 , … , i n + 1 = i + n + 1 i 1 = i , i 2 = i + 1 , … , i j = i + j − 1 i j + 1 = i + j + 1 , i j + 2 = i + j + 2 , … , i n + 1 = i + n + 1 {:[i_(1)=i","i_(2)=i+1","dots","i_(j)=i+j-1],[i_(j+1)=i+j+1","i_(j+2)=i+j+2","dots","i_(n+1)=i+n+1]:}\begin{gathered} i_{1}=i, i_{2}=i+1, \ldots, i_{j}=i+j-1 \\ i_{j+1}=i+j+1, i_{j+2}=i+j+2, \ldots, i_{n+1}=i+n+1 \end{gathered}i1=i,i2=i+1,…,iI=i+I−1iI+1=i+I+1,iI+2=i+I+2,…,in+1=i+n+1
and a simple calculation then gives us
(21)
[ x i , x i + 1 , … , x i + j − 1 , x i + j + 1 , … , x i + n + 1 ; f ] = = ( x i + j − x i ) Δ n i ( f ) + ( x i + n + 1 − x i + j ) Δ n i + 1 ( f ) x i + n + 1 − x i x i , x i + 1 , … , x i + j − 1 , x i + j + 1 , … , x i + n + 1 ; f = = x i + j − x i Δ n i ( f ) + x i + n + 1 − x i + j Δ n i + 1 ( f ) x i + n + 1 − x i {:[[x_(i),x_(i+1),dots,x_(i+j-1),x_(i+j+1),dots,x_(i+n+1);f]=],[=((x_(i+j)-x_(i))Delta_(n)^(i)(f)+(x_(i+n+1)-x_(i+j))Delta_(n)^(i+1)(f))/(x_(i+n+1)-x_(i))]:}\begin{gathered} {\left[x_{i}, x_{i+1}, \ldots, x_{i+j-1}, x_{i+j+1}, \ldots, x_{i+n+1} ; f\right]=} \\ =\frac{\left(x_{i+j}-x_{i}\right) \Delta_{n}^{i}(f)+\left(x_{i+n+1}-x_{i+j}\right) \Delta_{n}^{i+1}(f)}{x_{i+n+1}-x_{i}} \end{gathered}[xi,xi+1,…,xi+I−1,xi+I+1,…,xi+n+1;f]==(xi+I−xi)Δni(f)+(xi+n+1−xi+I)Δni+1(f)xi+n+1−xi
In general, for p = n + 1 , i j + 1 − i j = 2 p = n + 1 , i j + 1 − i j = 2 p=n+1,i_(j+1)-i_(j)=2p=n+1, i_{j+1}-i_{j}=2p=n+1,iI+1−iI=2we can write
[ x i 1 , x i 2 , … , x i n + 1 ; f ] = ( x i j + 1 − x i 1 ) Δ n i 1 ( f ) + ( x i n + 1 − x i j + 1 ) Δ n i 1 + 1 ( f ) x i n + 1 − x i 1 x i 1 , x i 2 , … , x i n + 1 ; f = x i j + 1 − x i 1 Δ n i 1 ( f ) + x i n + 1 − x i j + 1 Δ n i 1 + 1 ( f ) x i n + 1 − x i 1 [x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));f]=((x_(i_(j)+1)-x_(i_(1)))Delta_(n)^(i_(1))(f)+(x_(i_(n)+1)-x_(i_(j)+1))Delta_(n)^(i_(1)+1)(f))/(x_(i_(n+1))-x_(i_(1)))\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} ; f\right]=\frac{\left(x_{i_{j}+1}-x_{i_{1}}\right) \Delta_{n}^{i_{1}}(f)+\left(x_{i_{n}+1}-x_{i_{j}+1}\right) \Delta_{n}^{i_{1}+1}(f)}{x_{i_{n+1}}-x_{i_{1}}}[xi1,xi2,…,xin+1;f]=(xiI+1−xi1)Δni1(f)+(xin+1−xiI+1)Δni1+1(f)xin+1−xi1
Now suppose that the property is true for p = n p = n p=np=np=n, n + 1 , … , n + r n + 1 , … , n + r n+1,dots,n+rn+1, \ldots, n+rn+1,…,n+rand demonstrate that it will also be true for p = n + r + 1 p = n + r + 1 p=n+r+1p=n+r+1p=n+r+1.
If i n + 1 − i 1 = n + r + 1 i n + 1 − i 1 = n + r + 1 i_(n+1)-i_(1)=n+r+1i_{n+1}-i_{1}=n+r+1in+1−i1=n+r+1, formula (21) gives us
[ x i 1 , x i 2 , … , x i n + 1 ; f ] = = 1 x i n + 1 − x i 1 { ( x i j + 1 − x i 1 ) [ x i 1 , x i 2 , … , x i j , x i j + 1 , x i j + 1 , … , x i n ; f ] + + ( x i n + 1 − x i j + 1 ) [ x i 2 , x i 3 , … , x i j , x i j + 1 , x i j + 1 , … , x i n + 1 ; f ] x i 1 , x i 2 , … , x i n + 1 ; f = = 1 x i n + 1 − x i 1 x i j + 1 − x i 1 x i 1 , x i 2 , … , x i j , x i j + 1 , x i j + 1 , … , x i n ; f + + x i n + 1 − x i j + 1 x i 2 , x i 3 , … , x i j , x i j + 1 , x i j + 1 , … , x i n + 1 ; f {:[[x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));f]=],[=(1)/(x_(i_(n+1))-x_(i_(1))){(x_(i_(j+1))-x_(i_(1)))[x_(i_(1)),x_(i_(2)),dots,x_(i_(j)),x_(i_(j+1)),x_(i_(j+1)),dots,x_(i_(n));f]+:}],[+(x_(i_(n+1))-x_(i_(j+1)))[x_(i_(2)),x_(i_(3)),dots,x_(i_(j)),x_(i_(j+1)),x_(i_(j+1)),dots,x_(i_(n+1));f]]:}\begin{gathered} {\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} ; f\right]=} \\ =\frac{1}{x_{i_{n+1}}-x_{i_{1}}}\left\{\left(x_{i_{j+1}}-x_{i_{1}}\right)\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{j}}, x_{i_{j+1}}, x_{i_{j+1}}, \ldots, x_{i_{n}} ; f\right]+\right. \\ +\left(x_{i_{n+1}}-x_{i_{j+1}}\right)\left[x_{i_{2}}, x_{i_{3}}, \ldots, x_{i_{j}}, x_{i_{j+1}}, x_{i_{j+1}}, \ldots, x_{i_{n+1}} ; f\right] \end{gathered}[xi1,xi2,…,xin+1;f]==1xin+1−xi1{(xiI+1−xi1)[xi1,xi2,…,xiI,xiI+1,xiI+1,…,xin;f]++(xin+1−xiI+1)[xi2,xi3,…,xiI,xiI+1,xiI+1,…,xin+1;f]
Or j j jjIis an index such that i j + 1 < i j + 1 i j + 1 < i j + 1 i_(j)+1 < i_(j+1)i_{j}+1<i_{j+1}iI+1<iI+1. The property results immediately.
By taking f = x n f = x n f=x^(n)f=x^{n}f=xnin (20), we see that the sum of the coefficients in the second member is equal to 1. We therefore have the following property.
If the sequence (11) is ordered, any divided difference [ x i 1 , x i 2 , … , x i n + 1 ; f ] x i 1 , x i 2 , … , x i n + 1 ; f [x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));f]\left[\mathrm{x}_{i_{1}}, \mathrm{x}_{i_{2}}, \ldots, \mathrm{x}_{i_{n+1}} ; \mathrm{f}\right][xi1,xi2,…,xin+1;f]on n + 1 n + 1 n+1\mathrm{n}+1n+1of these points is an arithmetic (generalized) mean of the differences quoted Δ n 1 ( f ) , Δ n 2 ( f ) , … , Δ n m − n ( f ) Δ n 1 ( f ) , Δ n 2 ( f ) , … , Δ n m − n ( f ) Delta_(n)^(1)(f),Delta_(n)^(2)(f),dots,Delta_(n)^(m-n)(f)\Delta_{n}^{1}(\mathrm{f}), \Delta_{n}^{2}(\mathrm{f}), \ldots, \Delta_{n}^{m-n}(\mathrm{f})Δn1(f),Δn2(f),…,Δnm−n(f). So we have
[ x i 1 , x i 2 , … , x i n + 1 ; f ] = ∑ j = 1 m − n A j Δ n j ( f ) A j ≧ 0 , j = 1 , 2 , . . , m − n , ∑ j = 1 m − n A j = 1 x i 1 , x i 2 , … , x i n + 1 ; f = ∑ j = 1 m − n   A j Δ n j ( f ) A j ≧ 0 , j = 1 , 2 , . . , m − n , ∑ j = 1 m − n   A j = 1 {:[{:[x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));f]=sum_(j=1)^(m-n)A_(j)Delta_(n)^(j)(f):}],[A_(j) >= 0","j=1","2","..","m-n","sum_(j=1)^(m-n)A_(j)=1]:}\begin{aligned} & {\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} ; f\right]=\sum_{j=1}^{m-n} \mathrm{~A}_{j} \Delta_{n}^{j}(f)} \\ & \mathrm{A}_{j} \geqq 0, j=1,2, . ., m-n, \sum_{j=1}^{m-n} \mathrm{~A}_{j}=1 \end{aligned}[xi1,xi2,…,xin+1;f]=∑I=1m−n HASIΔnI(f)HASI≧0,I=1,2,..,m−n,∑I=1m−n HASI=1
THE A j A j A_(j)\mathrm{A}_{j}HASIbeing independent of the function f .
The coefficients A j A j A_(j)\mathrm{A}_{j}HASIare given by formula (20). The previous demonstration shows us, moreover, that
A j = 0 , j = 1 , 2 , … , i 1 − 1 , i n + 1 − n + 1 , i n + 1 − n + 2 , … , m − n , A j = 0 , j = 1 , 2 , … , i 1 − 1 , i n + 1 − n + 1 , i n + 1 − n + 2 , … , m − n , A_(j)=0,j=1,2,dots,i_(1)-1,i_(n+1)-n+1,i_(n+1)-n+2,dots,m-n,\mathrm{A}_{j}=0, j=1,2, \ldots, i_{1}-1, i_{n+1}-n+1, i_{n+1}-n+2, \ldots, m-n,HASI=0,I=1,2,…,i1−1,in+1−n+1,in+1−n+2,…,m−n,
which can also be seen in formula (20), taking into account definition (15) of the functions t j ∗ t j ∗ t_(j)^(**)t_{j}^{*}tI∗. The coefficients A i 1 , A i n + 1 − n A i 1 , A i n + 1 − n A_(i_(1)),A_(i_(n+1)-n)\mathrm{A}_{i_{1}}, \mathrm{~A}_{i_{n+1}-n}HASi1, HASin+1−nare surely positive. We can easily calculate their values. The function f i ∗ f i ∗ f_(i)^(**)f_{i}^{*}fi∗can be considered, on the points x i 1 , x i 2 … , x i n + 1 x i 1 , x i 2 … , x i n + 1 x_(i_(1)),x_(i_(2))dots,x_(i_(n+1))x_{i_{1}}, x_{i_{2}} \ldots, x_{i_{n+1}}xi1,xi2…,xin+1, as the sum of the polynomial f 1 + 1 , n − 1 f 1 + 1 , n − 1 _(f_(1)+1,n-1){ }_{f_{1}+1, n-1}f1+1,n−1of degree n − 1 n − 1 n-1n-1n−1and a function g = g ( x ) g = g ( x ) g=g(x)g=g(x)g=g(x)equal to φ i 1 + 1 , n − 1 ( x i 1 ) φ i 1 + 1 , n − 1 x i 1 varphi_(i_(1)+1,n-1)(x_(i_(1)))\varphi_{i_{1}+1, n-1}\left(x_{i_{1}}\right)φi1+1,n−1(xi1)For x = x i 1 x = x i 1 x=x_(i_(1))x=x_{i_{1}}x=xi1and zero outside this point. We then have
A i 1 = ( x i 1 + n − x i 1 ) [ x i 1 , x i 2 , … , x i n + 1 ; f i ∗ ] = ( x i 1 + n − x i 1 ) [ x i 1 , x i 2 , … , x i n + 1 ; g ] = = ( − 1 ) n + 1 ( x i 1 + n − x i 1 ) V ( x i 2 , x i 3 , … , x i n + 1 ) q i 1 + 1 , n − 1 ( x i 1 ) V ( x i 1 , x i 2 , … , x i n + 1 ) = = ( x i 1 + 1 − x i 1 ) ( x i 1 + 2 − x i 1 ) … ( x i 1 + n − x i 1 ) ( x i 2 − x i 1 ) ( x i 3 − x i 1 ) … ( x i n + 1 − x i 1 ) A i 1 = x i 1 + n − x i 1 x i 1 , x i 2 , … , x i n + 1 ; f i ∗ = x i 1 + n − x i 1 x i 1 , x i 2 , … , x i n + 1 ; g = = ( − 1 ) n + 1 x i 1 + n − x i 1 V x i 2 , x i 3 , … , x i n + 1 q i 1 + 1 , n − 1 x i 1 V x i 1 , x i 2 , … , x i n + 1 = = x i 1 + 1 − x i 1 x i 1 + 2 − x i 1 … x i 1 + n − x i 1 x i 2 − x i 1 x i 3 − x i 1 … x i n + 1 − x i 1 {:[A_(i_(1))=(x_(i_(1)+n)-x_(i_(1)))[x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));f_(i)^(**)]=(x_(i_(1)+n)-x_(i_(1)))[x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1));g]=],[=((-1)^(n+1)(x_(i_(1)+n)-x_(i_(1)))V(x_(i_(2)),x_(i_(3)),dots,x_(i_(n+1)))q_(i_(1)+1,n-1)(x_(i_(1))))/(V(x_(i_(1)),x_(i_(2)),dots,x_(i_(n+1))))=],[=((x_(i_(1)+1)-x_(i_(1)))(x_(i_(1)+2)-x_(i_(1)))dots(x_(i_(1)+n)-x_(i_(1))))/((x_(i_(2))-x_(i_(1)))(x_(i_(3))-x_(i_(1)))dots(x_(i_(n+1))-x_(i_(1))))]:}\begin{gathered} \mathrm{A}_{i_{1}}=\left(x_{i_{1}+n}-x_{i_{1}}\right)\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} ; f_{i}^{*}\right]=\left(x_{i_{1}+n}-x_{i_{1}}\right)\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}} ; \mathrm{g}\right]= \\ =\frac{(-1)^{n+1}\left(x_{i_{1}+n}-x_{i_{1}}\right) \mathrm{V}\left(x_{i_{2}}, x_{i_{3}}, \ldots, x_{i_{n+1}}\right) q_{i_{1}+1, n-1}\left(x_{i_{1}}\right)}{\mathrm{V}\left(x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+1}}\right)}= \\ =\frac{\left(x_{i_{1}+1}-x_{i_{1}}\right)\left(x_{i_{1}+2}-x_{i_{1}}\right) \ldots\left(x_{i_{1}+n}-x_{i_{1}}\right)}{\left(x_{i_{2}}-x_{i_{1}}\right)\left(x_{i_{3}}-x_{i_{1}}\right) \ldots\left(x_{i_{n+1}}-x_{i_{1}}\right)} \end{gathered}HASi1=(xi1+n−xi1)[xi1,xi2,…,xin+1;fi∗]=(xi1+n−xi1)[xi1,xi2,…,xin+1;g]==(−1)n+1(xi1+n−xi1)V(xi2,xi3,…,xin+1)qi1+1,n−1(xi1)V(xi1,xi2,…,xin+1)==(xi1+1−xi1)(xi1+2−xi1)…(xi1+n−xi1)(xi2−xi1)(xi3−xi1)…(xin+1−xi1)
We find in the same way
A i n + 1 − n = ( x i n + 1 − x i n + 1 − 1 ) ( x i n + 1 − x i n + 1 − 2 ) … ( x i n + 1 − x i n + 1 − n ) ( x i n + 1 − x i 1 ) ( x i n + 1 − x i 2 ) … ( x i n + 1 − x i n ) A i n + 1 − n = x i n + 1 − x i n + 1 − 1 x i n + 1 − x i n + 1 − 2 … x i n + 1 − x i n + 1 − n x i n + 1 − x i 1 x i n + 1 − x i 2 … x i n + 1 − x i n A_(i_(n+1)-n)=((x_(i_(n+1))-x_(i_(n+1)-1))(x_(i_(n+1))-x_(i_(n+1)-2))dots(x_(i_(n+1))-x_(i_(n+1)-n)))/((x_(i_(n+1))-x_(i_(1)))(x_(i_(n+1))-x_(i_(2)))dots(x_(i_(n+1))-x_(i_(n))))\mathrm{A}_{i_{n+1}-n}=\frac{\left(x_{i_{n+1}}-x_{i_{n+1}-1}\right)\left(x_{i_{n+1}}-x_{i_{n+1}-2}\right) \ldots\left(x_{i_{n+1}}-x_{i_{n+1}-n}\right)}{\left(x_{i_{n+1}}-x_{i_{1}}\right)\left(x_{i_{n+1}}-x_{i_{2}}\right) \ldots\left(x_{i_{n+1}}-x_{i_{n}}\right)}HASin+1−n=(xin+1−xin+1−1)(xin+1−xin+1−2)…(xin+1−xin+1−n)(xin+1−xi1)(xin+1−xi2)…(xin+1−xin)
Note. The previous demonstration shows us that the functions f j ∗ f j ∗ f_(j)^(**)f_{j}^{*}fI∗are non-concave of order n − 1 n − 1 n-1n-1n−1on points (11). This property can also be established directly.
6. The divided difference of a function of function. We have established formula (18) giving the divided difference of a product of two functions. Let us now consider a function of function G ( f ) G ( f ) G(f)\mathrm{G}(f)G(f)and we propose to find an expression for the divided difference
F = [ x 1 , x 2 , … , x n + 1 ; G ( f ) ] F = x 1 , x 2 , … , x n + 1 ; G ( f ) F=[x_(1),x_(2),dots,x_(n+1);G(f)]\mathrm{F}=\left[x_{1}, x_{2}, \ldots, x_{n+1} ; \mathrm{G}(f)\right]F=[x1,x2,…,xn+1;G(f)]
If we pose f i = f ( x i ) , i = 1 , 2 , … , n + 1 f i = f x i , i = 1 , 2 , … , n + 1 f_(i)=f(x_(i)),i=1,2,dots,n+1f_{i}=f\left(x_{i}\right), i=1,2, \ldots, n+1fi=f(xi),i=1,2,…,n+1, the expression F is linear and homogeneous in G ( f 1 ) , G ( f 2 ) , … , G ( f n + 1 ) G f 1 , G f 2 , … , G f n + 1 G(f_(1)),G(f_(2)),dots,G(f_(n+1))\mathrm{G}\left(f_{1}\right), \mathrm{G}\left(f_{2}\right), \ldots, \mathrm{G}\left(f_{n+1}\right)G(f1),G(f2),…,G(fn+1), therefore of the form
F = ∑ i = 1 n + 1 λ l G ( f i ) F = ∑ i = 1 n + 1   λ l G f i F=sum_(i=1)^(n+1)lambda_(l)G(f_(i))\mathrm{F}=\sum_{i=1}^{n+1} \lambda_{l} \mathrm{G}\left(f_{i}\right)F=∑i=1n+1λLG(fi)
So we will have
F = ∑ i = 1 n + 1 μ i [ f 1 , f 2 , … , f i ; G ] F = ∑ i = 1 n + 1   μ i f 1 , f 2 , … , f i ; G F=sum_(i=1)^(n+1)mu_(i)[f_(1),f_(2),dots,f_(i);G]\mathrm{F}=\sum_{i=1}^{n+1} \mu_{i}\left[f_{1}, f_{2}, \ldots, f_{i} ; \mathrm{G}\right]F=∑i=1n+1μi[f1,f2,…,fi;G]
We have
λ i = 1 j i 1 , n + 1 ′ ( x i ) , i = 1 , 2 … , n + 1 λ i = 1 j i 1 , n + 1 ′ x i , i = 1 , 2 … , n + 1 lambda_(i)=(1)/(j_(i1,n+1)^(')(x_(i))),i=1,2dots,n+1\lambda_{i}=\frac{1}{{\underset{i 1, n+1}{j}}^{\prime}\left(x_{i}\right)}, i=1,2 \ldots, n+1λi=1Ii1,n+1′(xi),i=1,2…,n+1
and formula (14) shows us that if we set
g j = g j ( x ) = ( f − f 1 ) ( f − f 2 ) … ( f − f j − 1 ) , g 1 = 1 j = 1 , 2 , … , n + 1 g j = g j ( x ) = f − f 1 f − f 2 … f − f j − 1 , g 1 = 1 j = 1 , 2 , … , n + 1 {:[g_(j)=g_(j)(x)=(f-f_(1))(f-f_(2))dots(f-f_(j-1))","quadg_(1)=1],[j=1","2","dots","n+1]:}\begin{gathered} g_{j}=g_{j}(x)=\left(f-f_{1}\right)\left(f-f_{2}\right) \ldots\left(f-f_{j-1}\right), \quad g_{1}=1 \\ j=1,2, \ldots, n+1 \end{gathered}gI=gI(x)=(f−f1)(f−f2)…(f−fI−1),g1=1I=1,2,…,n+1
We have
(22) μ j = [ x 1 , x 2 , . . x n + 1 ; g j ] , j = 1 , 2 , … , n + 1 (22) μ j = x 1 , x 2 , . . x n + 1 ; g j , j = 1 , 2 , … , n + 1 {:(22)mu_(j)=[x_(1),x_(2),..x_(n+1);g_(j)]","j=1","2","dots","n+1:}\begin{equation*} \mu_{j}=\left[x_{1}, x_{2}, . . x_{n+1} ; g_{j}\right], j=1,2, \ldots, n+1 \tag{22} \end{equation*}(22)μI=[x1,x2,..xn+1;gI],I=1,2,…,n+1
In particular, we have μ 1 = 0 μ 1 = 0 mu_(1)=0\mu_{1}=0μ1=0. Taking into account (18), formulas [ x 1 ; f − f 1 ] = 0 , [ x 2 ; f − f 2 ] = 0 x 1 ; f − f 1 = 0 , x 2 ; f − f 2 = 0 [x_(1);f-f_(1)]=0,[x_(2);f-f_(2)]=0\left[x_{1} ; f-f_{1}\right]=0,\left[x_{2} ; f-f_{2}\right]=0[x1;f−f1]=0,[x2;f−f2]=0and relations (7), (8), (9), we can write
μ 2 = ∑ i = 3 n + 1 [ x 1 , x 3 , … x i ; f ] [ x i , x i + 1 , … , x n + 1 , x 2 ; f ] μ 2 = ∑ i = 3 n + 1   x 1 , x 3 , … x i ; f x i , x i + 1 , … , x n + 1 , x 2 ; f mu_(2)=sum_(i=3)^(n+1)[x_(1),x_(3),dotsx_(i);f][x_(i),x_(i+1),dots,x_(n+1),x_(2);f]\mu_{2}=\sum_{i=3}^{n+1}\left[x_{1}, x_{3}, \ldots x_{i} ; f\right]\left[x_{i}, x_{i+1}, \ldots, x_{n+1}, x_{2} ; f\right]μ2=∑i=3n+1[x1,x3,…xi;f][xi,xi+1,…,xn+1,x2;f]
This coefficient is therefore a sum of n − 1 n − 1 n-1n-1n−1products of divided differences of order ≥ 1 ≥ 1 >= 1\geq 1≥1of f f fff, each product being formed by two divided differences whose sum of orders is equal to n n nnn. Let us demonstrate that in general
The coefficient μ j μ j mu_(j)\mu_{j}μIis a sum of products of divided differences of order ≧ 1 ≧ 1 >= 1\geqq 1≧1of f f f\mathfrak{f}f, each product being formed by divided differences whose sum of orders is equal to n.
Each divided difference is taken, of course, at certain points of the sequence x 1 , x 2 , … , x n + 1 x 1 , x 2 , … , x n + 1 x_(1),x_(2),dots,x_(n+1)x_{1}, x_{2}, \ldots, x_{n+1}x1,x2,…,xn+1
The demonstration is done by induction. We have seen that the property is true for the coefficient μ 2 μ 2 mu_(2)\mu_{2}μ2, regardless of n n nnn. Assume the property is true for the coefficient μ j μ j mu_(j)\mu_{j}μIfor all possible values ​​of n n nnnand let's demonstrate it for μ j + 1 μ j + 1 mu_(j+1)\mu_{j+1}μI+1. We have, taking into account (22), formula (18) and g j + 1 = g j ( f − f j ) g j + 1 = g j f − f j g_(j+1)=g_(j)(f-f_(j))g_{j+1}=g_{j}\left(f-f_{j}\right)gI+1=gI(f−fI),
(23) μ j + 1 = ∑ i = j + 1 n + 1 [ x 1 , x 2 , … , x j − 1 , x j + 1 , … , x i ; g j ] [ x i , x i + 1 , … , x n + 1 , x j ; f ] (23) μ j + 1 = ∑ i = j + 1 n + 1   x 1 , x 2 , … , x j − 1 , x j + 1 , … , x i ; g j x i , x i + 1 , … , x n + 1 , x j ; f {:(23)mu_(j+1)=sum_(i=j+1)^(n+1)[x_(1),x_(2),dots,x_(j-1),x_(j+1),dots,x_(i);g_(j)][x_(i),x_(i+1),dots,x_(n+1),x_(j);f]:}\begin{equation*} \mu_{j+1}=\sum_{i=j+1}^{n+1}\left[x_{1}, x_{2}, \ldots, x_{j-1}, x_{j+1}, \ldots, x_{i} ; g_{j}\right]\left[x_{i}, x_{i+1}, \ldots, x_{n+1}, x_{j} ; f\right] \tag{23} \end{equation*}(23)μI+1=∑i=I+1n+1[x1,x2,…,xI−1,xI+1,…,xi;gI][xi,xi+1,…,xn+1,xI;f]
which demonstrates the property.
This formula also tells us the number of terms of the coefficient μ j μ j mu_(j)\mu_{j}μI. Either N j n N j n N_(j)^(n)\mathrm{N}_{j}^{n}NInthe number of terms of μ j μ j mu_(j)\mu_{j}μIFor n + 1 n + 1 n+1n+1n+1points. We then have N 2 n = n − 1 N 2 n = n − 1 N_(2)^(n)=n-1\mathrm{N}_{2}^{n}=n-1N2n=n−1and formula (23) shows us that N j + 1 n = N j j − 1 + + N j j + … + N j n − 1 N j + 1 n = N j j − 1 + + N j j + … + N j n − 1 N_(j+1)^(n)=N_(j)^(j-1)++N_(j)^(j)+dots+N_(j)^(n-1)\mathrm{N}_{j+1}^{n}=\mathrm{N}_{j}^{j-1}+ +\mathrm{N}_{j}^{j}+\ldots+\mathrm{N}_{j}^{n-1}NI+1n=NII−1++NII+…+NIn−1from which we easily deduce,
N j n = ( n − 1 ) ( n − 2 ) ⋅ ( n − j + 1 ) ( j − 1 ) ! N j n = ( n − 1 ) ( n − 2 ) ⋅ ( n − j + 1 ) ( j − 1 ) ! N_(j)^(n)=((n-1)(n-2)*(n-j+1))/((j-1)!)\mathrm{N}_{j}^{n}=\frac{(n-1)(n-2) \cdot(n-j+1)}{(j-1)!}NIn=(n−1)(n−2)⋅(n−I+1)(I−1)!
Let us designate by d r ′ , d r ′ ′ , … d r ′ , d r ′ ′ , … d_(r)^('),d_(r)^(''),dotsd_{r}^{\prime}, d_{r}^{\prime \prime}, \ldotsdr′,dr′′,…divided differences of order r r rrrof f f ffftaken from groups of r + 1 r + 1 r+1r+1r+1points of the sequence x 1 , x 2 , … , x n + 1 x 1 , x 2 , … , x n + 1 x_(1),x_(2),dots,x_(n+1)x_{1}, x_{2}, \ldots, x_{n+1}x1,x2,…,xn+1. We then have
(24) μ j = ∑ d 1 ′ d 1 ′ ′ … d 1 ( j 1 ) d 2 ′ d 2 ′ ′ … d 2 ( j 2 ) … d n ′ d n ′ ′ … d n ( j n ) (24) μ j = ∑ d 1 ′ d 1 ′ ′ … d 1 j 1 d 2 ′ d 2 ′ ′ … d 2 j 2 … d n ′ d n ′ ′ … d n j n {:(24)mu_(j)=sumd_(1)^(')d_(1)^('')dotsd_(1)^((j_(1)))d_(2)^(')d_(2)^('')dotsd_(2)^((j_(2)))dotsd_(n)^(')d_(n)^('')dotsd_(n)^((j_(n))):}\begin{equation*} \mu_{j}=\sum d_{1}^{\prime} d_{1}^{\prime \prime} \ldots d_{1}^{\left(j_{1}\right)} d_{2}^{\prime} d_{2}^{\prime \prime} \ldots d_{2}^{\left(j_{2}\right)} \ldots d_{n}^{\prime} d_{n}^{\prime \prime} \ldots d_{n}^{\left(j_{n}\right)} \tag{24} \end{equation*}(24)μI=∑d1′d1′′…d1(I1)d2′d2′′…d2(I2)…dn′dn′′…dn(In)
Or
(25) j 1 + j 2 + … + j n = j , j 1 + 2 j 2 + … + n j n = n (25) j 1 + j 2 + … + j n = j , j 1 + 2 j 2 + … + n j n = n {:(25)j_(1)+j_(2)+dots+j_(n)=j","quadj_(1)+2j_(2)+dots+nj_(n)=n:}\begin{equation*} j_{1}+j_{2}+\ldots+j_{n}=j, \quad j_{1}+2 j_{2}+\ldots+n j_{n}=n \tag{25} \end{equation*}(25)I1+I2+…+In=I,I1+2I2+…+nIn=n
The summation (24) extends to all integer and positive or zero solutions of the system (25) with respect to j 1 , j 2 , … , j n j 1 , j 2 , … , j n j_(1),j_(2),dots,j_(n)j_{1}, j_{2}, \ldots, j_{n}I1,I2,…,In. Any solution corresponds to n ! j 1 ! j 2 ! … j n ! n ! j 1 ! j 2 ! … j n ! (n!)/(j_(1)!j_(2)!dotsj_(n)!)\frac{n!}{j_{1}!j_{2}!\ldots j_{n}!}n!I1!I2!…In!terms in μ j μ j mu_(j)\mu_{j}μI.
7. The case of differentiable functions. We have assumed that the points (11) are all distict. We can also assume the opposite. We then obtain limit formulas from (17) by making several of the points tend x i x i x_(i)x_{i}xitowards each other. The divided differences which are introduced then have the values ​​which we have given in our Thesis 6 6 ^(6){ }^{6}6). We
let us suppose, well extended, that these are now functions defined in an interval containing the points x i x i x_(i)x_{i}xi et un nombre suffisant de fois dérivables. Par exemple si tous les points viennent se confondre, la formule (18) devient la formule de Leibniz donnant la dérivée nème d'un produit. La formule de la différence divisée d'une fonction de fonction devient la formule donnant la n eme n eme  n^("eme ")n^{\text {eme }}neme  dérivée d'une fonction de fonction.
Bucureşti, le 29 octobre 1940.
  1. 1 1 ^(1){ }^{1}1 ) Tiberiu Popoviciu "Sur quelques propriétés des fonctions d'une ou de deux variables réelles" Thèse, Paris 1933 ou Mathematica, 8, 1-85 (1934).
    2 2 ^(2){ }^{2}2 ) Voir par ex. J. F. Steffensen „Note on divided differences" Danske Vid. Selsk. Math-fys. Medd. 17, Nr. 3, 1-12 (1939).
    3 3 ^(3){ }^{3}3 ) Un polynome de degré n n nnn est une expression de la forme
    c o x n + c 1 x n − 1 + … + c n c o x n + c 1 x n − 1 + … + c n c_(o)x^(n)+c_(1)x^(n-1)+dots+c_(n)c_{o} x^{n}+c_{1} x^{n-1}+\ldots+c_{n}coxn+c1xn−1+…+cn
    les c i c i c_(i)c_{i}ci étant des constantes positives, nulles ou négatives (il suffit de nous limiter au cas réel).
  2. 5 5 ^(5){ }^{5}5 ) En vertu de l'unicité ơu polynome (1). Nous tenons compte du fait qu'un polynome de degré n n nnn est complètement déterminé par la valeur de son premier coefficient et ses valeurs en n n nnn points distincts.
  3. 6 6 ^(6){ }^{6}6 ) Voir loc. cit. 1 1 ^(1){ }^{1}1 ), p. 43.
1940

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